# Physics FAQs and Topic Writeups

Discussion in 'MCAT Study Question Q&A' started by Shrike, Jun 15, 2005.

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1. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
This thread is a reference work, not a discussion. It comprises answers to Frequently Asked Questions (FAQs) about the physics portion of the MCAT, and write-ups of topics that give many students trouble on the test. Some of the posts are detailed, because the topics are non-trivial. Those in-depth posts are intended to guide your studying, not to substitute for other materials. We hope that this thread becomes a permanent resource for the SDN community.

It will be some time before this thread is completed. If you want to suggest a topic for inclusion (which request I will happily consider), please PM me. Meanwhile, readers and contributors should expect stuff to move around on a daily basis while construction continues.

Please do not post in this thread without permission. If you have permission, fire away, but be willing to have your posts edited, for format and content.

Thread moderated by: Shrike. Shrike is a full-time instructor for The Princeton Review; he has taken the MCAT twice for no good reason, scoring 14 on the physical sciences section each time. He majored in mathematics and economics, minored in physics, and spent several years accumulating unused school experience (in law).

**********

• Post 23: Acceleration, Speed, and Velocity
• Post 13: Applications of Thermodynamics (First Law of Thermodynamics, Calorimetry and Refrigerators--from Gen Chem Explanations Thread)
• Post 16: Buoyancy: Solving Problems
• Post 27: Capacitors
• Post 28: Capacitors: Examples
• Post 14: Centripetal and Centrifugal Forces
• Post 04: Conservation Laws
• Post 17: Conversion Units and Dimensional Analysis
• Post 19: Conversion Units II
• Post 22: Coulomb's Law
• Post 25: Electric Fields
• Post 26: Electric Potential and Work (Electrostatics)
• Post 09: How to Read Physical Science Passages
• Post 20: Neurotransmission from a Physics Perspective
• Post 10: Newton's Third Law
• Post 03: Optics Problems: How to Solve Them
• Post 21: Pendulums and Springs: Tension
• Post 15: Physics Constants Necessary for the MCAT and OAT
• Post 11: Physics Equations
• Post 12: Physics Equations II
• Post 13: Physics Equations III
• Post 08: Right Hand Rule for Magnetic Fields
• Post 18: Simple Machines
• Post 24: Solving Slope Problems
• Post 02: Solving Trig Functions without a Calculator
• Post 07: Tension: Ropes, Cables, Wires, Strings
• Post 05: Waves: Transverse and Longitudinal
• Post 06: Weightlessness

3. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
The easiest way to solve MCAT optics problems -- all of them -- is to know three equations (which must be memorized):

• 1/o + 1/i = 1/f
• m = -i/o
• p = 1/f
... nine easy definitions (try them; you already know most of them):

• o = distance (from the lens/mirror) to the object
• i = distance to the image
• f = focal length (sometimes hiding in the problem, as Radius of Curvature/2)
• m = magnification (which, oddly, includes being right side up and upside down)
• p = power
• converging = bringing together
• real = light goes there
• virtual = light does not go there
... three things you already know about the world, for translating MCAT problems and figures into the proper form:

• magnifying glasses, which are convex, make light come together
• mirrors are different from lenses (they have the opposite effect on light)
• concave is different from convex (it has the opposite effect on light)
... four easy, but perhaps novel, principles for setting up and interpreting the equations, based on the idea that positive is good:

• object o --> always + (because it's always real, and real is good)
• image i --> real is + (because real is good)
• focal length f --> converging is + (because it's good to come together)
• magnification m --> upright is + (because it's good to be upright)
... and one sensible method:

• to find the effect of mutiple lenses or mirrors, add their powers to get the power of the combination.

4. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
• In theory, momentum is always conserved, regardless of what happens. On MCAT problems, momentum is conserved if the system, i.e., all the objects that are mentioned, includes everything that exerts force on anything else. The most common trouble here is the Earth -- if gravity causes objects to fall, but the Earth is not mentioned in the problem, then AAMC would claim, confusingly, that momentum is not conserved.

Momentum and its conservation is usually mentioned in collision problems.

• Everything listed above for momentum is also true of angular momentum: it's always conserved in theory, but might not be according to AAMC if something is missing from the system. Angular momentum appears infrequently on the MCAT.

• Energy is not conserved, except in special cases: it is conserved during perfectly elastic collisions (you know that it's perfectly elastic because they tell you so), and when the only forces involved in a problem are "conservative."

Conservative forces on the MCAT are: gravity, electrostatic, springs; also, any physical pushing or pulling that doesn't involve dynamic friction. Dynamic friction, electrical resistance, radioactive decay, and any collision that is not perfectly elastic, are not conservative. If in doubt, it's not.

• Mass is conserved on the MCAT, except possibly in radioactive decay. If anyone has seen an exception to this principle in AAMC materials or tests, please advise.

• Net charge is conserved, always.

• The number of baryons (for us, protons and neutrons) is conserved, always. On the MCAT, we can just add them because we don't encounter antiprotons.

5. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
There are two kinds of waves, longitudinal and transverse. Longitudinal waves are waves that vibrate in the same direction that they travel (propagate). Transverse waves vibrate across their direction of travel.

Every physical material -- solid, liquid, and gas -- can conduct a longitudinal wave -- that's a compression wave, also known as sound (though often we call it sound only if it's of a frequency near the range of human hearing) -- in any direction. There are no longitudinal waves other than compression, i.e., sound, waves that matter on the MCAT.

Compression waves propagate in all directions through their media (i.e., sound diffuses).

Every physical material with a surface (i.e., every solid object, and every liquid/gas and liquid/liquid interface) can conduct a transverse wave in any direction along the surface, with displacement normal to the surface. Note that a thin wire is a special case, essentially a surface with normals pointing out of the wire in all directions, so it can vibrate in any direction perpendicular to it.

Though there have been MCAT passages with waves traveling in multiple directions on surfaces, these were unusual; in almost all problems, there is only one wave. Waves on surfaces, and especially liquid surfaces, can be very complex to model, so they don't appear very often.

Other waves are possible, including transverse waves far from the surface of objects. Though these waves might appear, their precise properties shouldn't matter on the MCAT.

In addition to waves in substances, light can be modeled as a transverse wave. Never mind what it's a wave of.

6. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
On the MCAT:

• Ropes, cables, wires, and strings are just ways to apply forces (except when they're vibrating...).

• The force on a rope etc. is always applied along the line of the rope, and it is the same in both directions. It is equal to, by definition, the tension on the rope.

• The amount of force, i.e., the tension, is the same throughout the rope. In particular, it is the same on each side of a pulley.

7. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
This is everything you need to know about the Right Hand Rule (RHR), or rather the two versions thereof. One or the other of them applies any time you are finding a direction (whether of field, force, acceleration, or even something else), and charges are moving.

I detail the flat hand method -- no sticking fingers in different directions perpendicular to each other.

There are two related right hand rules. Let's say: Right Hand Rule 1 (RHR1) -- used to determine the direction a charged particle, moving in a magnetic field, will be pushed; Right Hand Rule 2 (RHR2) -- used to determine the direction of a magnetic field created by the movement of charges.

There are three parts of your hand to remember, and I try to make it is easy as possible:

1. Fingers = Field: stick your fingers (of your right hand) in the direction of the magnetic field; if you aren't told the field (i.e., you are doing RHR2), the fingers are eventually going to tell you its direction. In other words, fingers = the first field you encounter in the problem. Remember, this is only for magnetic fields; electric fields have nothing to do with magnetism problems.

2. Thumb = hitchhike, in the direction of motion. Motion of what? Of the thing in the problem that's moving (the charged particle or particles; if there's just a current given, not particles, then that current would be the movement, by convention in the positive direction even though it's usually the movement of negative particles in the opposite direction). In other words, the first motion you encounter in the problem. (Note: the direction it's going now, not the direction of its acceleration.)

3. Palm = Push. Your palm will face in the direction that a positively-charged particle will be pushed by the magnetic field. For RHR1 problems, this is usually what the problem asked you to figure out. For RHR2 problems, it doesn't really apply but it still works: a positively-charged particle moving in the same direction as the current will be pushed toward the wire by the magnetic field.

So, if asked to figure out the direction a particle is pushed, stick your fingers in the direction of the magnetic field, and hitchhike in the direction the particle is moving. Your palm is now trying to push the charge, in the appropriate direction. You just used RHR1.

If asked to make a magnetic field, hitchhike in the direction of the moving charges (or the current); your fingers, which now are allowed to curl, show the direction of the field, which curls around the path of the charges. You just used RHR2.

Related issues:

• The magnetic field does zero work, always. This is because it is always perpendicular to motion, turning the particle rather than speeding it up or slowing it down. That's why we can kind of throw it in there at the end of electrostatics: it has no effect on the energy of anything.

• It is possible to use the analogous left hand rule to solve problems for negative charges moving in magnetic fields, but I recommend against it. Use RHR, of whichever type, and for negative charges or currents the answer is the opposite direction.

• Electric fields do not create magnetic fields; these two are unrelated for your purposes. You'll know this if you try to apply RHR inappropriately -- you can't stick your fingers in two different directions.

8. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
None. Or very close to none.

Time is of the essence on the Physical Sciences section; nearly everyone needs to finsh the section, but for most people that isn't going to happen if they spend too much time reading unnecessary material. And it turns out that most of the text in physics passage is unnecessary for answering the questions.

When you first turn to a physics passage, it is fine to glance at the first few lines to see what it's about, though even this isn't necessary. What you must do is check out every picture, table, graph, and block-quoted (offset) equation. You have a particular task with each of these items:

• Pictures: Often these show an experimental setup, or the system being studied. Examine it to see what's going on. If it's confusing, do not spend too much time here; sometimes, you don't really need to understand the entire experiment to answer all of the questions.

• Tables: These represent the most important part of many passages, because they show experimental results and AAMC loves experiments. For the moment (before you've read the questions), you have two tasks: (1) understand what the table is talking about, i.e., what experiment was performed -- you may have to read a few lines of text above the table in order to do this; and (2) see what was measured -- in other words, read and understand the column headings.

• Graphs: Do not analyze the graphs, just see what they are graphs of, by checking the labels on the axes and by reading a couple of lines of text above them if that's necessary.

• Equations: Do not attempt to understand any equation that is offset from the rest of the text, but do see what it's an equation for; again, this may necessitate reading a bit of text.

You then proceed to the questions, referring back to the pictures, tables, graphs, and equations as necessary.

About one question per passage, maybe not even that many, will require reading some more of the text, but by now you'll have a good idea where to look and what you're looking for, so you'll be able to find it quickly. Many people worry about these questions, but they're rarely a problem. If you have practiced enough passages and studied the right material, you'll know immediately when they are asking you something that requires reference to the passage, because it's not amazingly simple but it's also not something that you saw in the pictures, tables, graphs, or equations. For such not-so-simple questions, look in the passage.

You might think it would make sense to read the passage first if you're going to have to go back to it later. Not true. For one thing, some passages have no such questions. For another, for most test-takers reading the passage is a time sink. If you had the discipline to limit yourself to a quick skimming it would be fine, but under testing conditions most people hate to miss anything, so skimming becomes reading, and reading becomes time expiring before you're done or having to spend insufficient time on questions to compensate.

This method will be uncomfortable at first, and maybe forever. To those who are still reading the passages because they feel more comfortable when they do that, ask yourself whether the discomfiture is actually costing you points, and whether the time lost is worth it. If you're not 100 percent sure, test it both ways, but only after you've practiced the new method enough to get good at it. Remember, the point is getting questions right in the time allotted, not liking it.

9. ### xanthinesdecaying organic matterModerator Emeritus 10+ Year Member

Feb 22, 2002
In the classes I teach, a disconcerting number of students don't really understand Newton's third law of motion and get confused by the statement:

"For every action, there is an equal and opposite reaction."​

In fact, the law would be more accurately expressed this way:

"For every force, there is an equal and opposite force."​

Or, in mathematical terms:

"F1 on 2 = -F2 on 1"​

For a book lying on a table (i.e., stationary), it is commonly thought that Newton's 3rd Law is when the Normal Force and the weight of the book (m*g) cancel each other out and therefore the book goes nowhere. Actually, these two forces are not related by the Third Law. Given a (gravitational) force between the Earth and the book, the Third Law implies a force between the book and Earth, not the not the book and the table. Consider the math:

FG = G(MEarth)(Mbook)/ r^2​

You usually learn that acceleration due to gravity, g, is

g = GMEarth/r^2​

and the force due to gravity, aka weight, is mg. Newton's 3rd Law states that the book must exert the same amount of force on the Earth... and it does!

acceleration (a) of due to the book = G(Mbook)/r^2.​

Since in each case you're using masses of the book and earth for accelerations, the forces come out the same when you multiply by the mass of the other body. Forces are masses * accelerations, which means you will arrive at the same force whether or not you're using g or the acceleration due to the book. We've gotten used to just thinking about acceleration as g (which is immensely useful, no doubt), but it puts up blinders when thinking about forces, most notably weight.

Using the gravitational equation, you can figure out the force on the book due to the earth, as well as the force on the Earth due to the book. Each time, you will find the magnitude to be the same (although different directions). Using mg incorporates the mass of the book (or whatever) into the gravitational equation. Because of this, whenever you figure out the forces on the book or on the Earth, you end up using all the same variables. Try it out! This is what Newton meant by the Third Law.

-X

10. ### psiyung1K Member 10+ Year Member

2,421
24
Oct 23, 2004
These are some common Physics equations that will probably need to be memorized and understood for the MCAT

vectors

Pythagorean Theorem: a^2 + b^2 = c^2
**Remember the common 3-4-5 and 5-12-13 triangles**

sin(theta) = (opposite / hypotenuse)

Distance, Displacement, Velocity, and Acceleration

Displacement d = change in position
Speed = (distance) / (time)

Velocity v = (displacement) / (time) -- this is a vector
**Note that displacement is the NET distance from the starting point**

Acceleration a = (change in velocity) / (time) -- this is a vector
**Because acceleration is a vector, a change in velocity or a change in direction means the object has accelerated**

Linear Motion

v = v0 + at

d = v0t + 0.5at^2

vavg = 0.5(v + v0)

**Note that these equations are under the assumption that the objects are moving with constant acceleration**

Objects shot into projectile motion

dx = v0xt

V0x = V0xcos(theta) V0y = V0sin(theta)

ax = 0 ay = -10 m/s^2 (acceleration due to gravity)

Peak height of an object: V0 x sin(theta) = sqrt(2gh)

Range of object = Vicos(theta) x (time)

Newtons Law and the Law of Universal Gravitation

Force = mass x acceleration or F = ma

F = (Gm1m2) / r^2 where G = 6.67 x 10^-11 m^3 kg^-1 s^-2 (do not memorize this constant!)
**m1 and m2 are the masses of the objects while r is the distance between the center of masses of the two objects**

**Note that both objects described feel forces of the same magnitude**

**Note that combind these two equations we get**:
F = (Gm1m2) / r^2 = ma --> from here we can find the acceleration of either object just by plugging in the mass of the other object

Inclined Planes

F = mgsin(theta)
**this is the force causing an object to move up or down the incline plane**

Fn = mgcos(theta)
**this is the force being applied on the object that continuously changes the direction of velocity by creating a continuous centripetal acceleration**

Circular Motion

Centripetal Acceleration = ac = (v^2) / (r) *r = radius of circle*
**Note that this acceleration ALWAYS points toward the center of a circle**

Centripetal Force = F(c) = mv^2 / r

Frictional Forces

**Static friction -- Force opposing motion when two objects are not moving relative to each other**

f(s) = mus x Fn where mu(s) is the fraction of the Normal Force (Fn)

**Kinetic Friction -- Force opposing motion when two objects are moving relative to each other**

f(k) = mu(k) x F

**mu(s) > mu(k)**

Hooke's Law

F = -k(x)
k is a constant for a particular object
x is the deformation or the change in position of the object

Equilibrium

Means no linear or angular acceleration applied the an object (object moves and rotates at a constant velocity or zero velocity). The sum of all forces acting on such objects is zero

F(upward) = F(downward)
F(rightward) = F(leftward)

11. ### psiyung1K Member 10+ Year Member

2,421
24
Oct 23, 2004
Torque
Tau = (force) x (moment arm)

Tau(clockwise) = Tae (counterclockwise) for static problems

Center of mass = x = (m1x1+m2x2+....) / (m1+m2+....)
**Note the equation is dependent on the number of objects**

Energy

Kinetic Energy = K = .5(mv^2)

Gravitational Potential Energy = U(g) = mgh
**this equation is used for objects near the earth's surface**

Elastic Potential Energy = U(e) = .5(k)(x^2)
**this shows the potential energy for objects following Hookes Law**

Work

W = Fdcos(theta)
**Work is done by all forces except that of FRICTION**

W = (delta)K + (delta)U
**this equations can also be written:
.5(mv(f)^2) + mgh(f) = .5(mv(i)^2) + mgh(i)**

Power
P = (work) / (time) -- measured in watts
P = Fdcos(theta) / (time)
P = Fvcos(theta)
**sometimes power is defined by transferred energy (E) which is Work + heat(q)** P = (delta)E / (time)

momentum

p = mv -- mass and velocity of an object
**Momentum is always conserved**

Collisions

Elastic collisions: U(i) + K(i) = U(f) + K(f)
** Mechanical energies before and after collision are equal

Inelastic collision: Use the conservation of momentum to solve inelastic collision problems since mechanical energy is converted internal energy
**p(i) = p(f)

Impulse

Is the change in momentum: J = (delta)p

F(avg) = m(vf-vi) / (change in time)

mass defect
Nucleus of an atom has less mass than the sum of its individual parts. The difference of the mass is the mass defect. Binding energy holding protons and neutrons can be found using the defect

E = mc^2 where m is the mass defect

Density of and Specific Gravity

Density = mass / volume (kg/m^3)

S.G. = (density of a particular substance) / (density of water)

Fluid Pressure on objects

P = Force / area

P = (density)(g)
**where g is gravity and density and y is the density and depth of the fluid the object can be found respectively**

if a fluid is open to the atmosphere, the pressure is:
**Pressure = (density)(g) + P(atm)

Pascal's Principle

Any change in pressure applied to an enclosed fluid is transmitted undiminished to all parts of the fluid and the enclosing walls
**Principal seen in the Hydraulic Lift where: F(1)/A(1) = F(2)/A(2)

Archimedes Principle
Any fluid applies a buoyant force to an object that is partially or completely submerged in it -- this force is equal to the weight of the fluid that the object has displaced.

**F(b) = W(fluid)** -- can also be written

**F(b) = (density of fluid)(g)(Volume of fluid displaced)

Ideal Fluids

The equation of continuity: The mass flow rate of a fluid has the same value at every position along a tube that has a single entry and a single exit point for fluid flow --> for two positions along a tube:
**(density)(A1)(v1) = (density)(A2)(v2)**
A= cross sectional area of tube
v = fluid velocity

Since density is constant in an imcompressible fluid, it does not change during flow: A1v1 = A2v2
volume flow rate = Q = Av

Bernoulli's Equation
In a steady flow of nonviscous, imcompressible fluid of a cetain density, the pressure, the speed, and the elevation at two points are related by the following equation:

P1 + .5(density)(v1^2) + (density)(g)(y1) = P2 + .5(density)(v2^2) + (density)(g)(y2)

12. ### psiyung1K Member 10+ Year Member

2,421
24
Oct 23, 2004
Solids -- Included for completeness; DO NOT LEARN THESE! [according to Shrike]

Stress = (Force) / (area) -- Stress is what is done to an object
**The maximum stress applied to an object that allows the object to regain its original dimensions is called the yield point -- beyond this point, an object will not regain its shape**

Strain is the fractional change in the objects shape
**it equals the (change in dimension) / (original dimension)**

Always note that stress is proportional to strain:

The modulus of elasticity for a certain objects = (stress) / (strain)

Young's Modulus - deals with length changes in objects with an applied force
**F = (Y)(change in length)(Area) / (Original Length)**

Shear Modulus -- deals with shear deformation of objects with an applied force
**F = (S)(amount of shear)(Area) / (original length)**

Bulk Modulus -- deals with volume changes of objects by an applied pressure
** P = (-B)(change in volume) / (original volume)**
**Note that the negative sign denotes that anytime there is an increase in pressure, there is a decrease on volume**

Waves

V(wave) = (frequency)(wavelength)
**frequency is measured in hertz or cycles/second**

The period of a wave is the time required for one up and down cycle: On an x-axis of time, it is the point from any point of the wave function, to that exact point where the wave function has repeated itself
**T = (1/frequency)**

Intensity

Is measured in decibels -- &#946; = 10log(I/Ii)
**Note that for every factor of 10 that intensity increases, the decibels increase an additional 10 decibels**

beats
Beats occur when two waves of slightly different frequencies are superimposed. The beat frequncy produced will be the difference between the frequencies of the two original waves:

f(beat) = |f1 - f2|

Doppler Effect

fo = fs [(1)/(1 - v(s)/v)]
**This is for a source (s) moving towards a stationary observer**
**fo stands for the frequency head by the observer**
**v(s) is the speed of object the source is coming from while v is the speed of sounds (340 m/s)**

fo = fs [(1) / (1+v(s)/v) ]
**for a source moving away from an observer**

Electricity

F = (k)(q1)(q2) / ( r^2)
**Coulomb's Law -- remember that opposite charges attract each other and like charges repel each other**

Electric Field's due to a point charge

The electric field that exists at a point in space is the electrostatic force felt by a test charge at the point divided by the charge itself

E = (force) / (test charge)
**if we plug the Coulumbs equations we get E = k(q1)/(r^2)**
**test charge also shown as qo

Constant electrical fields

(force) = (qo)(E)

The electrical potential V at any given point is the electrical potential energy of a test charge qo, divided by the charge itself:

V = (EPE) / (qo)
**EPE (electrical potential energy) = (E)(q)(d)**
**plugging in these two equations we get V = Ed -- note that Voltage is the potential for work by an electric field to move any charge from one point to another**

Volatge by a point charge:
V = (k)(q1) / (r)

circuits

Voltage = IR (Ohm's Law)
R = resistance
I = current

Capacitance = Q / V Q = CV
Q = charge
V = voltage
**Capacitance is the ability to store charge per unit voltage**

The energy stored within a capacitor can be found using these equations:
U = .5(QV)
U = .5(C)(V^2)
U = .5(Q^2) / (C)

Resistors

Total resistance for resistors in series: R(eff) = R1 + R2 + R3 .......

Total resistance for resistance in parallel = 1/R(eff) = 1/R1 + 1/R2 + 1/R3......

Capacitors

Total capacitance of capacitors in series: 1/C(eff) = 1/C1 + 1/C2 + 1/C3 ...

Total capacitance of capacitors in parallel: C(eff) = C1 + C2 + C3 .....

Electric Power

P = IV
P = I^2(R)
P = V^2/R

magnetism

F = qvBsin(theta)
F = force on the charge moving through the magnetic field
q = charge
B = magnetic field
**Note that the force is directed perpendicular to both the velocity and the magnetic field**

Alternating Circuits

V(max) = sqrt(2) rms
I(max) = sqrt(2) rms

Light and Optics

Index of refraction

n = c/v
**Compares the speed of light in a vacuum with the speed of light through a given medium**

angle of refraction -- Snell's Law

n1sin(theta) = n2sin(theta)

E = hf
**this equation shows the higher the freqeuncies of light have more energy**

Mirrors and lenses -- learn the optics laws.

13. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
Not many. And those you need, you do not need with much precision at all.

• pi = 3 (Not kidding; this should be good enough. 3.1 in a pinch)
• sin(30) = 0.5; cos(30) = 0.9
• sin(45) = 0.7; cos(45) = 0.7
• sin(60) = 0.9; cos(60) = 0.5
• the speed of light in a vacuum (and air): c = 3 x 10^8m/s
• acceleration due to gravity near the surface of the Earth: g = 10 m/s^2
• the density of water, in the following units: 1000kg/m^3 = 1g/cm^3
Also useful:
• the speed of sound in air = 340 m/s
• 1 atmosphere = 100,000Pa = 10m of water = 760mmHg
• the density of water, in the following units: 1kg/L

14. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
First, you must understand this principle: the buoyant force, which is directed upward, is equal by definition to the weight of the fluid displaced.

Problem type 1
: the object is floating, in equilibrium, on the surface of a fluid.

If the density of the fluid above the surface can be neglected (i.e., it's much lighter than the fluid below, usually a gas above a liquid), then you need to understand two things:
• The buoyant force, which is equal to the weight, mg, of the fluid displaced, also equals the weight of the object (because the object is in equilibrium); it follows that the mass of the object is exactly equal to the mass of the fluid displaced.
• The portion of the object that is below the fluid equals the ratio of the object's density to that of the fluid.
The second principle follows from the first. Notably, it means that the way an object floats does not depend on how strong gravity is.

If the density of the fluid above the surface cannot be neglected (i.e., it's a liquid or a very dense gas), then the buoyant force is the sum of the weights of the respective fluids displaced by the portions of the object above and below the surface. In practice, subtract the density of the lighter fluid from the density of the the heavier fluid, use the principle above for solving problems neglecting the upper fluid to get a density of the object, and then add the density of the lighter fluid back in to get the virtual density of the object. (If you're finding how much is below the surface of the heavier fluid, subtract the density of the lighter one from both the heavier fluid and the object, and find the ratio as above.) An example shows it better:

Question: An object is floating at the surface of two fluids; two thirds of its volume is below the surface. If the densities of the two fluids are 1.0 g/cm^3 and 1.6g/cm^3, what is the density of the object?

Answer: Subtract the density of the lighter fluid from that of the heavier, and we get 0.6 g/cm^3. Two thirds of the object is below the surface, so that yields a density of 2/3 * 0.6 = 0.4 g/cm^3. Then we need to add the density of the first fluid back in: 0.4 + 1.0 = 1.4 g/cm^3.​

Problem type 2
: the object is below the surface of the fluid.

The easiest way to do these problems quickly is to do everything in terms of water: figure out what the buoyant force would be if the fluid were water, and then convert at the end. You need to have the density of water memorized in three different units: kg/m^3, g/cm^3, and kg/L. Note that a liter (L) is equal to 1000 cc's; many MCAT objects are most easily considered in these units.

To solve the problem, you will want to find the buoyant force, which is the weight of the fluid dispaced. To do this, you find the mass of the fluid displaced, and at first you pretend it's water. After finding the buoyant force, you subtract from the weight of the object to find its apparent weight; use that to get acceleration if needed. An example:

Question: A 20 cm x 40 cm x 5 cm object of density 0.4g/cm^3 is submerged with specific gravity of 1.2g/cm^3. Find the apparent weight of the object, and its acceleration.

Answer: The object's volume is 4000cm^3, or 4L. If the fluid were water, the mass displaced would therefore be 4kg (because the density of water is 1kg/L), so the buoyant force would be 4kg x g = 40N. This fluid is 1.2 times as dense as water, so the buoyant force is 1.2 x 40N = 48N.

The object is 0.4 times as dense as water, so its weight is 40N x 0.4 = 16N (and its mass is therefore 1.6 kg; we use that below). (Note that again we do everything in terms of water, because it's the easiest way. Note also that the volume of the object is, by definition, the same as that of the fluid displaced.) The apparent weight is therefore 16N - 48N = -32N. The negative weight means the net force on the object is upward; this makes sense because it's lighter than the fluid, i.e., it wants to float.

To find the object's acceleration, use F = ma --> a = F/m = 32N/1.6kg = 20m/s^2.​

The reliance on water makes many calculations easier to do quickly, and also provides a running sanity check -- you know it immediately when you are getting a nonsensical result -- but it actually doesn't change the method in a technical sense.​

15. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
The only conversion you will have to do is of prefixes, e.g., micro-whatsis to kilo-whatsis. No kilograms to slugs, or liters to hogsheads, or meters to chains, or...

For those who have trouble converting units, I suggest the following method:

First, forget the conversion factors listed in whatever reference material you're using. They're correct, of course, but if you're having trouble they're what're mixing you up. Instead, just think in terms of big and small, as shown here. Also, eschew multiplying by negative exponents -- divide instead. Finally, think of the units as variables, and cancel as you would with variables.

An example problem will show the method best. Lets say you want to know the speed of light, c = 3x10^8 m/s, but in cm/ms.

Write down 3x10^8 m/s on the left. Write down the units you want, cm/ms, on the right. Now multiple and divide to get rid of the m and the s, and to get cm and ms in the right spots:

Code:
```3x10^8 [U]m[/U] x _____[U]cm[/U] x _____[U]s[/U]_ = _______[U]cm[/U]
s        m         ms          ms```

Now fill in the numbers, which you should know, just focusing on which unit is bigger, and remembering that it takes more of the smaller ones to make a big one:

Code:
```3x10^8 [U]m[/U] x [U]10^2[/U]_[U]cm[/U] x ___[U]1[/U]_[U]s[/U]_ = _______[U]cm[/U]
s      1 m    10^3 ms          ms```

Check to ensure that the old units cancel out, that you wind up with the new units, that you have more of the smaller units in each fraction (the fractions must each equal 1); now do the math:

Code:
```3x10^8 [U]m[/U] x [U]10^2[/U]_[U]cm[/U] x ___[U]1[/U]_[U]s[/U]_ = [U]3x10^7[/U]_[U]cm[/U]
s      1 m    10^3 ms          ms```

Finally, never use a squared or cubed conversion factor: use the normal conversion factor, and square or cube it at the end. As another example, find the density of osmium in kg/m^3:

Code:
```[U]22.4 g[/U]___ x ___[U]1[/U]_[U]kg[/U] x (____[U]cm[/U])^3 = _______[U]kg[/U]_
cm^3        g    (    m )^3          m^3```

Code:
```[U]22.4 g[/U]___ x ___[U]1[/U]_[U]kg[/U] x ([U]100[/U]_[U]cm[/U])^3 = [U]3x10^7[/U]_[U]kg[/U]_
cm^3   1000 g    (  1 m )^3          m^3```

Code:
```[U]22.4 g[/U]___ x ___[U]1[/U]_[U]kg[/U] x [U]10^6[/U]_[U]cm^3[/U] = [U]2.24x10^4[/U]_[U]kg[/U]_
cm^3   1000 g       1 m^3              m^3```

Notice that we filled in the cm/m conversion, which we know well, and squared at the end. Notice also that there's no need to use scientific notation for many numbers -- the notation is a tool, not an end in itself.

If this method doesn't solve your conversion problems, I'll be surprised.

16. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
Recent MCATs have included several questions about simple machines -- devices that alter the strength of the input force and the distance through which force is applied. You should be familiar with the force multiplication concept, and somewhat familiar with some of the machines that use it.

The basic concept is this: you can change force, but you can't get something for nothing, i.e., energy is not created or destroyed by any of these machines. The energy of a force is, more properly stated, the work done by that force; it follows that the work done by an input force is the same as that done on the object that is being moved (unless there's friction, in which case energy is lost; we'll neglect friction for the rest of this discussion). As work = force x distance x cos(theta), if we increase force we have to decrease distance equivalently, and vice versa. (For this discussion we'll consider forces that do work in the direction in which they are applied, i.e., cos(theta) = 1, and W = Fd.) The amount by which force is increased by a system, and therefore distance decreased, is the force multiplier.

Because of the fixed relationship between force and distance, it is often easier to consider distance (which we can see) than force, and then work backwards. For example, if you can see that the thing you're moving travels, say, three times farther than the force that moves it, then the force on the object is three times less than the input; if you can see that it travels only a quarter as far, then the force exerted on it is four times as great, and so on.

The simple machines (and other devices with force multipliers) follow, including all six classic simple machines (even those that aren't likely to appear in the test; I note which these are). Do not memorize these; instead, be moderately familiar with them, especially the more common ones.

• Pulley: A single pulley or a system thereof is not necessarily a simple machine, but it can be. For MCAT-style pulley systems, output force is input force multiplied by the number of ropes that are effectively pulling on the object to be moved; equivalently, by the number of ropes that must (not just might) shorten in order for the resistance to move. Another hint: if none of the pulleys is free to move, then the force multiplier is 1: force and distance remain the same. Of course, if the force multiplier is not 1, i.e., you get greater force on the object than you put in, you will get less distance by an equivalent factor. Appears on the MCAT.

• Inclined plane: the force multiplier (relative to straight lifting) is the sine of the angle of the plane with the horizontal. Distance is divided by the same factor. Consider the problem another way: If you push a block up an incline, you push along the hypotenuse, but lift only the amount of its height; the ratio of these two is sin(theta) Because distance is less than what you input, force must be greater by the same factor -- the force multiplier. Common on the MCAT.

• Lever: the force multiplier is the ratio of distance from fulcrum to object, to distance from fulcrum to applied force. (The "fulcrum" of a lever is its pivot point.) Look at the picture to make sure you know whether distance goes up or down (hence force down or up). This treatment works even when the force and the object are on the same side of the fulcrum. I saw a lever -- a claw hammer pulling a nail -- and a question about its force multiplier and another about its fulcrum, on the August 2004 MCAT, so you can't ignore this one.

• Hydraulic jack: not usually considered to be a simple machine like the aforementioned, but effectively the same. Force multiplier is the ratio of the areas of the tops of the two pistons. Recall that force changes, not pressure -- pressure is the same (at the same height) everywhere in a body of liquid. Assuming that the change in height is negligible, it also doesn't matter what liquid is used. Appears on the MCAT.

• Wedge: similar to an inclined plane; the force multiplier is length divided by width; distance is divided by the same factor. Unlikely on the MCAT.

• Wheel and axle: the force multiplier is ratio of radii of the wheel and the axle. To figure whether to divide or multiply the two, look at the picture to see whether the distance the object travels is greater or less than the distance over which the force is applied; that will tell you whether the force is decreased or increased, respectively. Unlikely on the MCAT.

• Gears (technically not a separate type of simple machine, but they work the same as the others): the force multiplier is the ratio of the numbers of teeth on the two wheels. Again, look at the picture to see which increases, force or distance. Unlikely on the MCAT.

• Screw: an inclined plane wrapped around a rod. Unlikely on the MCAT. For completeness: the force multiplier is 2 x pi x radius x (turns/unit length). Don't worry about this one at all.

It is possible to combine two or more of the types: for example, a cam is an inclined plane combined with a wheel and axle. But don't worry: such a problem is very unlikely to appear on the MCAT, and if it did it would necessarily be a very simple setup.

17. ### psiyung1K Member 10+ Year Member

2,421
24
Oct 23, 2004
Some people still have problems converting cm^3 and mm^3 to m^3 for problems such as density and volume.

1) When converting from cm^3 to m^3 move 6 decimal places to the left. For example if the volume of a box is 120 cm^3, then converting it, we will get .000120 m^3 or 1.2 x 10^-4 m^3

2) When converting from mm^3 to m^3 move 9 decimal places to the left. For example, if the volume is 120 mm^3, the conveting it, we will get .000000120 m^3.

**REASONING**

1 m^3 = 1m x 1m x 1m = 100 cm x 100 cm x 100cm = 1000mm x 1000mm x 1000 mm

18. ### QofQuimicaSeriously, dude, I think you're overreacting....Lifetime Donor SDN Administrator 10+ Year Member

Coulomb's Law describes what happens when two charges are placed close to one another. The magnitude of the force between the two charges, called the electrostatic force, depends on two variables: the magnitudes of the two charges, and the distance between the two charges. Whether the force is attractive or repulsive depends on the signs of the two charges: like charges will repel, and unlike charges will attract. Mathematically, two charges that repel will have a positive value for the electrostatic force, while two charges that attract will have a negative value for the electrostatic force.

Dependence on the Magnitudes of the Charges:
The electrostatic force is directly dependent on the magnitude of each of the two charges. If one charge's magnitude is doubled, the electrostatic force between them will be doubled. If the second charge's magnitude is doubled, again, this will double the force between them. If the magnitudes of both charges are doubled, the force between them will be quadrupled.

F q1
F q2

Dependence on the Distance Between the Charges:
The electrostatic force is indirectly dependent on the square of the distance between the two charges. This is why Coulomb's Law is known as an inverse square law, like Newton's Law of Universal Gravitation. If the distance between the two charges is doubled, the force is cut by one-fourth. If the distance between the two charges is halved, the force between them is quadrupled.

F 1/(r^2)

Putting all of these variables together with a permittivity constant gives us Coulomb's Law:

F = k(q1)(q2)/(r^2)

You should understand and memorize Coulomb's Law, but you do NOT need to memorize the value of k for the MCAT.

19. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
The good news is that you don't have to be able to do this on actual MCAT questions. All you would need to be able to do at the very most on a problem is to estimate where your trig function value will fall with relation to the five angles that you know. (You should know the cos and sin values for 0, 30, 45, 60, and 90 degrees; if you don't, memorize them NOW.) That'll be enough to choose the answer using process of elimination.

Here's the general approach to MCAT questions that ask for an angle as an answer. First, the only angles you need to know anything about are 30, 45, and 60 degrees (plus things like 0 and 90, plus equivalent angles like 135 [which is the same as 45]). Moreover, you don't need to know them with any precision: on MCAT physics questions, one significant figure is plenty. This means that for 30 degrees, sine is 0.5 and cosine is 0.9; for 60 degrees these two are reversed, and for 45 degrees the sine and cosine are each 0.7. (Warning: MCAT chemistry questions sometimes do require more than one significant figure, but there's no trig needed for them.)

Note about tangent: I don't know when you'd need tangent on the MCAT. To sum, you only need to know the following angles for sine and cosine:
• cos (0) = 1; sin(0) = 0
• cos(30) = 0.9; sin(30) = 0.5
• cos(45) = 0.7; sin(45) = 0.7
• cos(60) = 0.5; sin(60) = 0.9
• cos(90) = 0; sin(90) = 1
Any other angle, you will be given the trig functions for, or you will only have to see that the function is bigger/smaller than one of the angles listed above. The mnemonic for trig functions is SOH CAH TOA: sine = opposite over hypotenuse, cosine = adjacent over hypotenuse, tangent = opposite over adjacent.

20. ### ShrikeLanius examinatianus 10+ Year Member

646
4
Apr 23, 2004
too far south
Shrike: Weight is defined as the force of gravity on an object, and always equals mg. In popular parlance, an object is "weightless" when its apparent weight is zero. Apparent weight is just what a scale would read if the object were sitting on it; i.e., it's the force exterted on it, in an upward direction, by what it's resting on (or, occasionally, hanging from). An object is therefore weightless when the floor et al is not pushing it up; in other words, when it's in free fall.

Another example of apperent weight being zero is a neutrally buoyant object submerged in a fluid, or any object floating in equilibrium on the surface of the fluid. Because the buoyant force balances the force of gravity, no more force is needed to support it and a scale under the object would read zero.​

Xanthines: Weight is the force due to gravity, which on earth is mg. An object has no weight when there is no acceleration due to gravity. Alternatively, I guess an object could be considered weightless if something cancels out mg, like objects with densities identical to water that are submerged in water. This is why NASA conducts some of its training underwater and why senior citizens excercise in water. They don't have to deal with the forces being exerted by that pesky gravity.​

21. ### xanthinesdecaying organic matterModerator Emeritus 10+ Year Member

Feb 22, 2002
Xanthines: Here is an explanation of centripetal force, using the example of your car making donuts in a level parking lot. In this case, the engine is providing the power for forward motion (by forward, I mean the direction of motion of the front of the car). The centripetal force equation Fc=(m)(v^2)/r describes the force required to keep the object in a circular path. The static friction force points towards the center of the circle, and this is the source of the centripetal force that keeps the car moving in its circular path for this example:

Fc = uFn = mv^2/r​
Without a centripetal force, objects such as cars and yo-yos on strings will move in a direction that is tangential to the circular path they were on previously. Basically, what happens is that when you let go of the steering wheel, the car will move in a straight line that is tangential to the donuts you were making before. The car is not actually being pushed away from the center of the circle; rather, the centripetal force had been holding it in its circular motion, and now that centripetal force is gone. In addition, if the car moves fast enough, the force due to static friction will not be high enough to maintain circular motion. Again, that is because on level surfaces, the centripetal force is the friction force...I've heard the term "centripetal force requirement" used to describe this situation. This example shows us why it is that without friction, there can be no circular motion for cars. For example, when you try to make turns on icy patches of the road, you find that you cannot turn in a circle. There is no friction to provide a centripetal force, and thus you skid off the road into a ditch. Note that this is only true for level surfaces. On banked highways and race tracks, gravity comes into play.

To sum: centripetal force isn't really a new type of force, but rather is just the name used to indicate the net force pointing inwards for an object in circular motion. In the case of a car making donuts, the centripetal force is the static friction force. Since the static coefficient of friction is normally given as the maximum, you know that any more force applied to that object will overcome the friction, and the object will begin to slide in a direction tangent to the circle of motion (or in the case of the car, skid off the road.) Analogously, when a satellite moves fast enough, it will escape the earth's gravity:

G(Mearth)(Msatellite)/(r^2) is less than (msatellite)(v^2)/r​
One final point: When most people use the term "centrifugal force," they are actually referring to what is commonly referred to by another physics term known as inertia.

Shrike: It's true that by "centrifugal force," non-physicists usually mean something else: either inertia, or centripetal force. But to clarify what Xanthines said, there *is* such a thing as centrifugal force. Recall that, by Newton's Third Law, for every force there's an equal and opposite force, and that the two constitute an action-reaction pair. Centrifugal force is the force that forms a pair with the centripetal force -- it's the force exerted by the body that's moving in a circle, on whatever's making it turn.

For example, in the case of a yo-yo whirled on a string, the force of the string on the yoyo is centripetal, and the force of the yo-yo on the string, and thus your finger, is centrifugal. In the case of the car driving in a circle, the ground exerts a centripetal foce on the tires, while the tires exert a centrifugal force on the ground. You might guess from these examples that it's centripetal force that we usually worry about, not centrifugal. You'd be right. Centrifugal force tends to be a little odd, and mostly irrelevant. I've never seen it matter on an MCAT problem.

22. ### NutmegValar Morghulis 10+ Year Member

27,097
4,233
Aug 18, 2003
Grasnia
QofQuimica: You may recall from your neurobiology lessons that the axons of neurons are myelinated, and that myelin speeds the transmission of action potentials. However, you may not have ever thought about why this is true. This post explains an application of neurotransmission concepts to biology (cell membranes). So for all of you who are wondering why you have to study physics, and what physics has to do with anything in the "real world" for a pre-health student, here is one great example. It will hopefully also increase your intuitive understanding of the principles behind myelination and neurotransmission.

Lindyhopper: My remembrance is that myelin increases the resistance of the membrane, and therefore reduces its capacitance. If we compare a myelinated axon with an unmyelinated axon, the unmyelinated axon has a larger capacitance, so more charge must be deposited on the membrane to change the potential across the membrane. Thus, the current must flow for a longer time to produce a given depolarization. I'm trying to come up with a quick, simple analogy that might possible be helpful to explain this. I was thinking, perhaps, of comparing the unmyelinated axon to a soaker hose & the myelinated to a normal garden hose. But this has problems. The ions charging the capacitor do not normally flow through the membrane. I'm also not sure if the water flowing through the soaker hose flows significantly more slowly.

Nutmeg: If you're looking for an anology to understand membrane capacitance, here's the one that I always use for wave propagation. Imagine that you have one huge, heavy, chain and one long, unassembled fiberglass tent pole (the kind you use for a dome tent, where you have lengths about a foot long, connected by a single elastic cord that runs through the whole length) of equal length. The tent pole will sag a bit, yes, but the chain sags completely. Now imagine trying to hold one end and getting the other end to respond to a quick snap--basically, you're trying to whip it. With the chain, it will take enormous effort to get any degree of propagation because there are so many places where the movement gets transferred. With the tent pole, each segment is relatively stiff, and movement of one section sends (propagates) the energy much further than does a single chain link.

I usually think of these sorts of matters in terms of stiffness when trying to propagate a wave. You end up with a continuum, from a heavy steel chain, to a cotton jump rope, to a stiff nylon rope, to a bamboo pole, to a steel rod. The things that matter are the number of joints (more jointed things will be slower and take more nergy), the weight (the heavier it is, the more energy it needs), and the stiffness. In the case of the axon, the myelin makes the axon have fewer "joints." So, instead of a constant in-out flux down the entire length of the axon membrane, the local depolarization at the nodes of Ranvier can cause brief ionic diffusion along the length of the axon only at discrete points. The "weight" in this situation is analogous to the amount of charge required to propagate the signal. The decreased capacitance of the myelinated axon means that less charge is needed to propagate an action potential, as Lindyhopper stated. Likewise, the heavy chain needs more work to get the same pulse size as the light chain.

Finally, there is the stiffness. In the world of ropes and rods, a stiff region will behave very similarly to neighboring regions. Something floppy, however, can easily bend, and it doesn't transfer energy as quickly or efficiently. The internodal regions, in this example, are very "stiff". They do not have local variations because they can't communicate with the extracellular space very well. Thus, when one area changes in polarity, since the membrane cannot pass ions to or from the extracellular space, the adjacent intracellular regions will pass ions instead. You've reduced what are known as the degrees of freedom--the number of possible modes that can be adopted by the axon--such that charge will go up or down the length of the axon, but not in and out of the cell.

Or, if all that was too complicated, imagine a chain of people playing telephone. In each case, the message will be sent a distance of one kilometer. If you place the people at a distance of two meters apart and have them relay a message, one to the next, it will take significantly longer to send a message one kilometer than it would if you had spaced the people 25 meters apart. Analogously, having fewer nodes means that the message will be transmitted over the same distance in a shorter amount of time.

23. ### NutmegValar Morghulis 10+ Year Member

27,097
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Aug 18, 2003
Grasnia
It may be counterintuitive, but in a pendulum, the tension in the string during an oscillation is larger than mg. Here is an example to explain why this is so. Imagine for a moment that you are looking at a weight on a string that's bouncing, and imagine that you are also looking at a pendulum swinging, but that you're looking at it from the side, so that it's swinging toward and away from you. Now let's imagine that the string length, the masses of the weights, and the spring constant, etc., are all chosen so that the period of the oscillation is the same for both, and that the maximum and minimum heights of each weight is the same. From your perspective, they should look the same. Each weight should rise and fall in lock-step.

Now, when the spring-ball is at the bottom, it is easy to think that the tension on the spring must be greater than mg. If the tension were only mg, then the ball would sit motionless at the bottom. The added potential energy comes from the stretching of the spring, and this potential energy will convert to kinetic energy in lifting the ball up again. When it gets to the top, the potential energy comes from its height. This is how an oscillation works: conservation of energy, which converts from one form, to another, then back. In the case of the spring, the energy converts from potential (height) into kinetic (falling down) to potential (of the spring tension) to kinetic (the ball springing back up) and back to the start. This is a bit easier to understand intuitively than the pendulum, but it's the same principle.

With the pendulum, let's imagine that the center of rotation is the center of a clock, where the pendulum swings from eight o'clock, down to six, up to four, and then reverses back to six, and up to eight. At eight o'clock and four o'clock, the ball has no kinetic energy--it pauses motionless for an infinitessimally short moment. All of its energy is potential, from the height. At six o'clock, there is no potential energy. All of its energy is kinetic, as it reaches its maximum speed. At the time that the weight is at the bottom or the arc, the direction of the velocity is perpendicular to its arc. The string has the job of redirecting that kinetic energy to make the ball go up to either 8 or 4 o'clock. To do this, the direction of acceleration is perpendicular to the direction of the motion. This is centripetal force. Now, if you imagine a ball hanging motionless from a string, it has no kinetic energy. The total tension on the string in this situation is just -mg, as it is at equilibrium. The swinging pendulum at the bottom of its arc is not at equilibrium at all. Specifically, it has the same force of gravity as the ball at rest, but since the ball in the pendulum is swinging, it also has the force created by the acceleration--that is, the change in direction of the velocity--that comes at the bottom.

I think the key here is to remember that acceleration is not just the change in the speed, but the change in velocity. When you first hear about speed and velocity, it seems like a nit-pick to say that they are not the same, and that velocity is speed + direction. But a point on a wheel spinning at constant speed still accelerates. When you see that F = m*a, remember that acceleration doesn't care if it's a change in speed or a change in direction--it creates a force all the same. Since the pendulum ball changes direction of velocity, it must necessarily be accelerating, and since it has mass, there must be a resultant force. If the only tension on the string when the ball was at the bottom was -mg, then the system would be at equilibium, and the ball would be completely motionless.

24. ### Dave_DSenior Member 7+ Year Member

515
0
Jun 19, 2005
Massachusetts
Solving slope problems.

Let's start simply with a frictionless inclined plane and build from there, since the steps are largely the same for any block on an inclined plane like the one pictured here:

Getting the forces, probably the hardest part
The first part of solving an inclined plane problem is figuring out all of the forces acting on the block. We'll start by looking at gravity, which has 2 components. The first is the component of gravity that tries to slide the block down the plane, called the sliding force (F=mg sin x). The second is the component of gravity forcing the block against the plane, called the y component of the block's weight (F=mg cos x). Since I never could remember which component was sin and which was cos, I'll give you an easy way to just figure it out. Think about if the angle was 0 (i.e., the plane was flat). Then there's no force pushing the block down the plane. Now, which trig function gives us 0 with an angle of 0? It's sine. Likewise, the force pushing down is the full force of gravity; which trig function gives us 1 with an angle of 0? That would be cosine.

Friction, it justs rubs you the wrong way.
Ok, so now you've got the force sliding the block down the plane and the force pushing it against the plane; what next? Well, you need to consider, is there friction? If not, you're done; lucky you. The net force is just the sliding force; you can use F = ma to get the acceleration, mass, or whatever they're asking for.

So if you're not lucky, they put in friction. The next question is whether the force is enough to overcome static friction. Maybe they'll tell you that the block is in fact moving or has just started to move. Cool, in that case, static friction has been breached; you get to figure out dynamic friction, and you can skip to the next section. Of course, these problems are rarely that simple, and you'll probably have to work through static friction. The good news is that it's not that much more work. Static friction seems complex given that the formula is F < or = uN, where u is the coefficient of static friction and N is the normal force. The normal force is the force that the incline plane pushes up against the block. (It is very convenient to keep the block from sinking into the inclined plane that the normal force does this. ) Anyway, N is equal in magnitude to the force of gravity pushing the block down into the plane, i.e., F = mg cos x, the y-component to the weight of the block. However, N points in the opposite direction of the y component of the weight.

So what's the deal with the inequality anyway? What static friction tries to do is generate exactly enough force to resist motion. So if we had a 10N force sliding the block down the plane and static friction could generate anything up to 100N to resist this motion, it would actually generate a 10N force in the opposite direction and your block will just sit there. However, you just need to calculate the maximum force of static friction by using the formula F = uN. (Oh, and if you used signs and got a negative number, make the result positive, or you'll probably confuse yourself.) So ask yourself at this point, "Is the force sliding this block down the plane greater than the maximum force of static friction?" If you answered no, then the prof threw you a break and the block isn't going anywhere. Otherwise, we can now forget that we ever figured out static friction and go to the next section.

Dynamic (Kinetic) Friction
We have good news: no inequalities are needed for dynamic friction, and the formula is exactly the same:

Fdynamic friction = uN

Ok, if your block is moving, and friction is present, you need to subtract the friction force opposing the motion from the x component of the gravitational force. This gives you the net force. If we call the x component of the gravitational force the sliding force, we would get the following equation for the net force:

Fnet = Fsliding - Fdynamic friction

Oh, if somehow you got a negative number at this point, you definitely screwed something up, since negative forces in this context mean that the block is sliding up the plane for no real reason. (Of course, if you had a couple of rocket engines strapped to the block, maybe it really IS sliding up the plane, but somehow I doubt that.)

25. ### dbhvtSenior Member 5+ Year Member

Aug 9, 2005
SPEED vs VELOCITY

As far as speed is concerned, remember that speed is a scalar (just a number) and velocity is a vector (a number and a direction). You could come up with a situation where there is no change in speed but a change in velocity. That situation would be a particle moving in a circle without speeding up or slowing down (i.e., uniform circular motion). The velocity will be changing because the particle is not moving in a straight line. Even though the numerical value of the velocity isn't changing, the direction part is changing, and therefore the velocity is also changing.

ACCELERATION, VELOCITY, AND SPEED

Acceleration is defined as a change in velocity over some time.

So if the net force is not zero, then every single time without fail no matter what, the acceleration is not zero and the velocity is changing.

If the net force is constant, the acceleartion is constant and the velocity is changing at a constant rate.

Say you have a constant acceleration of 10 m/s^2 down.

Every second, you add 10 m/s down to your previous velocity vector. If you start out going 5 m/s down, in 1 second you'll be going 15 m/s down, in two you'll be going 25 m/s down, etc....

If you have non-zero acceleration, but it is NOT constant, your velocity vector will change, but it will change different amounts every second. Eg, start out at 5 m/s down, next second your at 40 m/s down, the next second you're at 2 m/s to the left.

Again, if the acceleration is constant, and it is not zero, there IS a change in velocity. Always, every single time, no matter what. It could be a change in the direction, the speed, or both. On the MCAT, acceleration problems will generally be constant non-zero acceleration. All of those problems will involve a change in velocity. Everyone, everytime, every year, every MCAT administration. Now repeat after me:

"Acceleration IS a change in velocity. If there is acceleration, the velocity is changing."

HINT:

Try to stop thinking about speed. Speed doesn't really make sense. If you're going 50 mph, do you care what direction you are going? Yes, of course you do. Doing your thing around town do you ever care about speed without direction? No.

Speed is an abstract silly thing that is used when bragging about the Porsche they give you if you get a 45 on the MCAT. In physics and real life, you always deal with direction at the same time, so you want to think about velocity.

26. ### gridironModerator Emeritus 2+ Year Member

479
3
Jun 4, 2006
I think I will tackle the question on electric fields, electric potential and work done in an electric field in parts. First, I will discuss electric fields.

What is an electric field? Imagine you have a positive point charge and place a test positive charge in its vicinity. Coulomb's law tells us that there is repulsion between the two charges. How exactly do the two charges feel one another? Then answer is the positive point charge sets up an electric field in the space around it. At any point within the space, the electric field has both a magnitude and a direction. The magnitude is dependent on the value of the charge and direction depends on the electrical charge. The electric field is a vector field and is defined for each point in space. By definition, the electrostatic force a test charge qo will feel in the vicinity of a point charge is:

E = F/ qo

The unit for electric field is N/C.

How do we draw electric field lines? Electric field lines give us a representation of the electric field. Say we have a sphere of uniform negative electric charge. If a positive test charge were placed in the field, the test charge would feel an electrostatic force pointed toward the sphere. This tells us that for negative charges, electric field lines point in. What if we have a sphere of uniform positive electric charge? If a positive test charge were placed in the field, the test charge would feel an electrostatic force pointed away from the sphere. Therefore, for positive charges, electric field lines point outward. How do electric field lines relate to electric field vectors? Simple! The electric field vector is tangent to the electric field line. For straight field lines, this means the electric field is that direction!

What is the electric field due to a point charge? Instead of a uniform sphere of charge, suppose we have a point charge and place a test charge qo in the vicinity. How would we calculate the electric field? From above, we know E = F/ qo. F is the electrostatic force which is given by Kq qo/r2, where r is the distance between the charges. Using the equation above, we see this reduces to: Kq/r2. Now, for some important points:
1. The above equation gives us the value for the electric field for all points in space for a point charge
2. The electric field is a vector---VERY IMPORTANT!!!
3. If there is more than one point charge, the electric fields are additive.

What would happen if we were to place a test charge in an electric field produced by a moving charge? Well, the electrostatic force that the test charge would feel is given by:

F=qE

Where q is the electric charge of the particle, and E is the electric field produced by the moving charge. Examining this equation, we see that if a negative charge is placed in the electric field produced by a positive charge, there is a negative electrostatic force due to attraction (This attraction can be verified using Coulomb's law). Therefore, the direction of the electrostatic force has the direction of the electric field if the test charge is positive and will have the opposite direction if the test charge is negative. This is important! If on the MCAT, you are given a diagram of a electric field and given a test charge, and are asked the direction of the electrostatic force, you can use the procedure outlined above. Given all this information, you can also calculate the acceleration of the test charge. If the only force acting on the test charge is the electrostatic force, than you can set that value equal to F=ma, where F is the sum of the net forces. This can all be related to Newton's second law!!

Question: (imagine this was a MCAT question) What if you are given a diagram of a negative test charge in a electric field pointing outward and you asked to determine whether the acceleration would increase if the test charge were moving in the direction of the electric field? First, if the electric field is pointing outward, than the point charge is positive. From Coulomb's law, we know there is attraction between the test charge and point charge. Therefore, the electrostatic force, the net force, should be pointing inward. However, the test charge is moving away from the point charge. Since the electric field decreases with distance (Kq/r2), this means if qE=ma, than the acceleration will decrease. That's it! On the MCAT, you would circle the answer choice with decreases!

Now, for a real world application of electric fields and charge. Believe it or not, you can use this analysis for ink-jet printers. You don't need to know this for the MCAT!!!

In the printer, when the print command is received, a generator generates drops. These drops then pass through a charge unit and receive a charge q. These drops then pass through a pair of conducting plates where the electric field points down from the top plate to the bottom plate. Once these drops pass through the plates, they are deflected land on the paper. By varying the magnitude of the electric field and charge of the drop, you get drops that land at different points on the paper. When this is done thousands of times, you get a printed copy of your work!!! (FYI: This is how older versions of ink-jet printers operate. There is much more to how they operate, but this is the basics!!) Who knows, a passage such as this might be on the MCAT!!!

Up Next: Electric potential (Coming 6/24)

27. ### gridironModerator Emeritus 2+ Year Member

479
3
Jun 4, 2006
What is electric potential? By definition, electric potential is the potential energy per unit charge. The unit for electric potential is the volt (V). According to the definition of electric potential, the unit can also be expressed as (J/C). This is very important because the MCAT may use both interchangeably. So, what exactly does electric potential mean? The best way to explain electric potential is to use the concepts that govern the macroscopic world and to apply them to the microscopic world of charges. Objects possess some charge that can be positive, negative or neutral. The charge which constitutes the objects sets up an electric field around the object that acts as a detector. This electric field is what allows the object to detect neighboring objects or fields. If the field points outward than the object has a positive net charge and if the field points inward the object has a negative net charge. If another charged object were to enter the field, an electric force would act on the object accelerating it in some direction. Which direction? The direction of the net force acting on the charged objectthis is what Newtons second law tells us. But, how do you determine direction? Imagine that you have a positive source charge. The direction of the electric field for the source charge will be outward. Now, you place a positive test charge in the electric field. What happens? The direction of the net force acting on the test charge is outward since it will repel the source charge. Thus, it will accelerate in the direction of the field. The opposite will happen with a negative test chargeit will accelerate in the direction opposite of the electric field. You can calculate the magnitude of the force using the following equation:

F=qE

Where q is the test charge and E is the magnitude of the electric field of the source charge.

The electric field is a vector approach to explain the behavior of charged objects (similar to velocity and moving objects). Electric potential is a scalar approach to explain the behavior of charged objects (similar to kinetic energy and moving objects). From Newtons second law, if an object moves in the direction of the net force acting on it, the velocity of the object will increase. This means the kinetic energy will increase. Conversely, if the object moves opposite the direction of the net force acting on it, the kinetic energy will decrease. For charged objects, the electric potential explains the energy of an object with respect to its position in some electric field. So, how do you explain the potential and kinetic energy of a charged object? Consider the macroscopic world. If you were to lift a book from the ground over your head, what kind of energy does the book gain? It gains potential energy. If you were to then drop the book from above your head, what happens? The potential energy translates into kinetic energy. When you lift the book over your head, you are doing something the book would not normally do. When you drop the book from the top of your head, the book is doing what it wants to doto move in the direction of gravity and to the floor. This means that when an object does something it naturally will not do, it will gain potential energy. When an object does something it naturally wants to do, it will gain kinetic energy (This makes sense according to Newtons first law). This same idea can be applied to the microscopic world. When a positive test charges moves in the direction of the electric field due to a positive source charge, the test charge is doing something it naturally wants to do, potential energy is lost and kinetic energy is gained. This means for the system to gain potential energy some external agent needs to act on the system to push the two like charges near one another. Only then will potential energy be gained.

In a nutshell:

• When two like charges repel, this is a natural tendency. Thus potential energy is lost and kinetic energy is gained.
• When opposite charges attract, this is a natural tendency. Thus potential energy is lost and kinetic energy is gained.
• In order to move two like charges near one another, this will not naturally happen because they will repel, external work needs to be done on the system. This means potential energy is gained and kinetic energy is lost.

So now that the concepts are explained, let me introduce some equations. The mathematical representation for electric potential is:

V = U/q

Where V is the electric potential, U is the potential energy and q is the charge.

The potential energy of a test charge is:

U = qV.
This will be negative when potential energy is lost and positive when potential energy is gained. This is a very important point for the MCAT because when asked to calculate the potential energy of the system, you can knock out choices you know are wrong based on what you know of the tendency of the system. This saves time on the test and prevents unnecessary calculations. Work is defined as:

W = -&#916;U

Thus negative work means a positive change in potential energy. This is another important point for the MCAT.

I hope this covers the important points! If you have any questions or concerns please post in the physics thread. Good .

28. ### gridironModerator Emeritus 2+ Year Member

479
3
Jun 4, 2006
There are many different ways to store potential energy. You can stretch a spring, compress a gas or lift an object over your head. You can also use an electric field to store potential energy and a capacitor does just that. Capacitors have many uses in electronic devices. Examples include computers and the defibrillator. Capacitors also come in different shapesfrom the parallel plate capacitor to even spherical and cylindrical. For the MCAT, the parallel plate capacitor is most important.

The standard parallel plate capacitor can be described by two parallel plates separated by some distance. The plates have area A and the distance between them is d. When an external agent, like a battery, charges a capacitor the plates acquire equal but opposite charge. To calculate the charge (q) acquired on any plate:

q = CV

Where C is the capacitance and V is the voltage. The voltage is the potential difference between the plates. The MCAT likes proportions so try the following question:

Q: The charge is doubled but the potential difference between the plates is kept constant. What happens to the capacitance?

A: Since this is a proportion, and q = C, then the capacitance will double. WRONG!! The capacitance stays the same!

The value of the capacitance is dependent on the geometry of the capacitor and is independent of the charge and potential difference (voltage). For a parallel plate capacitor the capacitance (C) is equal to:

C = &#1108;A/d

Where &#1108; is the permittivity constant, A is the area of the plate and d is the distance between the plates. Gauss law is needed to derive this equation (surprisingly Gauss law is a topic on the AAMC list of topics for the MCAT). Dont worry about the value of the permittivity constant because it will be provided on the MCAT if it necessarywhich is very unlikely because you are more likely to encounter a proportion question. From the equation if the area of the plates is increased the capacitance will increase but if the distance between the plates is increased the capacitance will decrease. A larger area means more room for charge to be storedmore charge means more energy. But, what about the distancewhy does an increase mean a decrease in capacitance? That will be addressed in a bit!

What if you put capacitors in parallel or series?

Case 1: Capacitors in parallel arrangement.

When capacitors are connected in a parallel fashion to a battery source, all the capacitors are at the same potential. This potential is the voltage of the battery source. How do you calculate the equivalent capacitance for two capacitors, C1 and C2, in parallel? For resistors in parallel, the equivalent resistance is 1/Req = 1/R1+..1/Rn. That is not true for capacitors. For capacitors in parallel, the potential difference is the same across each. Therefore:

q1 = C1V and q2 = C2V​

The total charge is then:

q = q1 + q2 = (C1 + C2)V

C1 + C2 represents the equivalent capacitance, Ceq, so:

q = CeqV -------- Ceq = q/V = C1 + C2

Thus, for capacitors arranged in a parallel fashion, the equivalent capacitance is found by:

Ceq = C1 + C2 + .. Cn (where n is the nth capacitor). Thats it! This is very important for the MCAT!​

Case 2: Capacitors in series arrangement.

When capacitors are arranged in series to a battery, the total potential difference across all the capacitors is equal to the voltage of the battery. This means that all capacitors end up with the same charge, q. How is this? Consider the case where you have two capacitors, C1 and C2, connected in series to a battery source. The negative terminal of the battery is connected to C2 and the positive terminal to C1. This means that the bottom plate of C2 will acquire a negative charge. As the bottom plate acquires negative charge, this will repel negative charge on the top plate of C2. Due to repulsion, the top plate of C2 will acquire a positive charge. The repelled negative charge will move to the bottom plate of C1. The charge repelled will be the magnitude of the charge on C2. Thus, both capacitors end up having the same charge. How do you find the equivalent capacitance? The potential difference across each capacitor:

V1 = q/C1 V2 = q/C2 ------ q is the same for both​

V = V1 + V2 = q ( 1/C1 + 1/C2) -------- (1/C1 + 1/C2) = Ceq

Ceq = q/V = 1/(1/C1 + 1/C2)

Thus,

1/Ceq = 1/C1 + 1/C2

For capacitors in series, the equivalent capacitance is then found by:

1/Ceq = 1/C1 +1/Cn (where n is the nth capacitor in series) ----very important for the MCAT!​

So, how exactly is energy stored in a capacitor? As charge builds up on the plates of the capacitor, equal but opposite charge, work is being done by an external agent to charge the capacitor. As equal but opposite charge builds up, an electric field builds up between the plates of the capacitor. Remember, positive and negative charges want to attract one another, but in order to separate the two opposite charges external work is needed. For capacitors, this is most often done by a power source such as a battery. Since opposite charges are being separated by one another, potential energy is being stored within the electric field. Work is done by the battery over the distance of the plates in order to store this potential energy in the electric field. This is why as the distance between the plates of a parallel plate capacitor is increased, the capacitance, the ability to store charge, decreases. More work needs to be done by the battery as charge builds up on the plates and over time it becomes more difficult to charge the capacitor. Over time, the potential energy stored in the capacitor is:

U = 0.5CV^2------very important for the MCAT!​

What if you put some material between the plates of the capacitor? What happens to its ability to store charge? This is the case of a dielectric material. Will the capacitance increase as a result of a dielectric? The answer is YES! For the MCAT you dont need to understand why but it is the result of the electric field between the plates decreasing. Since the electric field is less between the plates due to a dielectric, more charge can be stored and the capacitance increases. The dielectric, depending on the material, is some constant denoted by: &#1179;. For a parallel plate capacitor with a dielectric, the capacitance is given by:

C = &#1108; &#1179; A/d------the capacitance increases by the dielectric constant!!

29. ### gridironModerator Emeritus 2+ Year Member

479
3
Jun 4, 2006
Q: What would happen to the capacitance value, C, for a capacitor if the charge, q, were doubled?

A: The MCAT loves proportions, but this is a trick question. Possible answer choices could be:

a. double
b. triple
c. no change
d. cannot be determined

Without a good understanding of capacitance, many might choose a, but that is the incorrect answer. Although q and C are directly proportional, capacitance is dependent on the geometry of the capacitor. Thus, the correct answer would be c.

Q: Suppose you have two capacitors C1 = 1F and C2 = 2F. C1 is charged to a voltage of 5 V by a battery. The battery is disconnected from C1 and C1 is then connected directly to C2. What will be the potential across each capacitor?

A: This question is a bit more involved. When C1 is connected to C2, C1 will begin to discharge. The total charge on C1 will begin to disperse between the two capacitors. After all, where else can the charge on C1 go? At equilibrium, the potential differences across each capacitor will be the same. Knowing this you can solve the problem.

qtot = q1 + q2

C1Vinitial = C1V + C2V

V = VinitialC1/C1 + C2

V = 5/3 volts or about 1.67 volts

It is important to note that charge cannot be destroyedthus the initial charge should equal the final charge.

Q: Suppose you have a capacitor connected to a battery and introduce a dielectric during charging. What happens to the potential difference and capacitance?

A: The capacitance always increases if a dielectric is introduced regardless of whether the battery is still connected. But what happens to the potential difference? It stays the same. How can it increase or decrease if it is still connected to the battery? It will be at the same potential as the battery.