Physics Question Thread 2

[Pembleton]

Hi guys,

I don't have my physics textbook nearby and I was wondering if someone could remind me how to figure out the center of mass of two objects on a plank, for instance.

For example a 10 kg box is 1 m from the edge of a 10 m plank and a 20 kg box is 2 m from the other edge of the plank. Where is the center of mass on the plank?

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[axp107]

A meteor of mass M collides w/ a planet of mass M/2 that is initially at rest. The meteor exerts a force of F on the planet, causing the planet to accelerate. At the same time, the planet exerts a force on the meteor of:

A) F/2
B) F
C) 2F
D) 0

I picked F/2 b/c .. I felt the a smaller force had to be exerted on the larger planet.. in order for the planet to accelerate

F - F/2 > 0.. so it will accelerate

except, the answer is F
why is this so (I know theres always an equal and opposite force exerted..) but if this is true.. how can the planet accelerate?

 

[gridiron]

A meteor of mass M collides w/ a planet of mass M/2 that is initially at rest. The meteor exerts a force of F on the planet, causing the planet to accelerate. At the same time, the planet exerts a force on the meteor of:

A) F/2
B) F
C) 2F
D) 0

I picked F/2 b/c .. I felt the a smaller force had to be exerted on the larger planet.. in order for the planet to accelerate

F - F/2 > 0.. so it will accelerate

except, the answer is F
why is this so (I know theres always an equal and opposite force exerted..) but if this is true.. how can the planet accelerate?

Hey! The answer is F because of Newton's third law. Newton's third law will tell you about the force but nothing about the acceleration. The forces will be equal and opposite but you need Newton's second law for the resulting acceleration. Even thought the forces acting on both objects are the same, the larger planet will accelerate less because a = F/m (where m is the mass). The key point is that both planets will have some acceleration and Newton's third law tells you nothing about acceleration. I hope this helps and good

 

[gridiron]

A meteor of mass M collides w/ a planet of mass M/2 that is initially at rest. The meteor exerts a force of F on the planet, causing the planet to accelerate. At the same time, the planet exerts a force on the meteor of:

A) F/2
B) F
C) 2F
D) 0

I picked F/2 b/c .. I felt the a smaller force had to be exerted on the larger planet.. in order for the planet to accelerate

F - F/2 > 0.. so it will accelerate

except, the answer is F
why is this so (I know theres always an equal and opposite force exerted..) but if this is true.. how can the planet accelerate?

Hey! The answer is F because of Newton's third law. Newton's third law will tell you about the force but nothing about the acceleration. The forces will be equal and opposite but you need Newton's second law for the resulting acceleration. Even thought the forces acting on both objects are the same, the larger planet will accelerate less because a = F/m (where m is the mass). The key point is that both planets will have some acceleration and Newton's third law tells you nothing about acceleration. I hope this helps and good .

 

[axp107]

Another question...

I hope I can explain it properly: You'll prolly have to draw a little diagram

Imagine a typical circuit with a + and - battery...

On the plus side of the plate, as you go along the wire, there is an ANODE
On the mius side of the plate, as you go along the wire, there is a CATHODE

Assume some outside presence (light for instance) knocks an electron off from the cathode, the electron moves towards the anode... it accelerates towards the anode.. why??

I thought electric field always points from + to -, cathode to anode... so if the electric field points in the + to - direction, since the charge is an electron, not a proton, won't it want to move in the OPPOSITE direction, thus slowing down as its knocked off?

OR

Do you look at the battery and say that current travels from +ive to the -ive battery terminal, so the electric field actually points from the anode to cathode.. if this is the case, what do anode and cathode really mean? whats the point of having them if all you have to do is look at the battery and which way current is travelling.

 

[tennisboy85]

Another question...

I hope I can explain it properly: You'll prolly have to draw a little diagram

Imagine a typical circuit with a + and - battery...

On the plus side of the plate, as you go along the wire, there is an ANODE
On the mius side of the plate, as you go along the wire, there is a CATHODE

Assume some outside presence (light for instance) knocks an electron off from the cathode, the electron moves towards the anode... it accelerates towards the anode.. why??

I thought electric field always points from + to -, cathode to anode... so if the electric field points in the + to - direction, since the charge is an electron, not a proton, won't it want to move in the OPPOSITE direction, thus slowing down as its knocked off?

Hey Axp,
I think you shouldn't always think of circuits in one dimension. So to review, in circuits, charges flow from Anode to Cathode, but this is "by convention" and similarly, "by convention" means the flow of positive charges. What actually happens in a real circuits is electrons (not positive charges) that flow - which would technically mean the flow in circuits is cathode to anode (or Negative to Positive)

Now, recall the "An Ox, Red Cat"
Anode = site of oxidation (aka losing e-)
Cathode = site of reducation (aka gaining e-)

If you have an electron flowing towards the Anode, as you mention, you can imagine that electrons would just LOVE to go there. Why? Because there are less e- there, therefore less negative charge.

OR

Do you look at the battery and say that current travels from +ive to the -ive battery terminal, so the electric field actually points from the anode to cathode.. if this is the case, what do anode and cathode really mean? whats the point of having them if all you have to do is look at the battery and which way current is travelling.

While I'm not 100% sure on the directions of current (because I think AC current switches their anode/cathode sites, thus the flow direction), electric fields "by convention" goes from positive to negative. So, in terms of battery, the +ve site should be the Anode (again, An Ox, oxidation = less electrons) and the -ve site should be Cathode (Red Cat).

The point of the word anode and cathode is just to confuse you. lol, jokes.

I hope this helps. I hope I didn't make a mistake there.

 

[jochi1543]

Guys, my Physics midterm is in 2 weeks, I'm already studying, and already freaking out. I've been doing some practice problems, but unfortunately, there are no answers to them in the book, so I was wondering if someone could evaluate my solutions, as there are 2 or 3 probs I'm not very confident about.

1) The electric field between 2 circular plates of a capacitor is changing at a rate of 3 x 10^6 V/m over 1 sec. If the displacement current at this instance is Id = 1.60 x 10^-8 A, find the dimensions of the plates.

My solution: Id = e delta F(electric flux)/ delta t, where e is the permeability constant (8.85...). So 1.60 x 10^-8 = 8.85x10^12 x delta F/1, if we consider a time of 1 second. So then delta F = (1.60x10^-8)/(8.85x10^-12) = 1808 Vm/s.

Then I said, F(flux) = EA, so delta F = delta E x A. We need to find the area, so A = delta F/delta E, which is 1808 Vm/s divided by 3 x 10^6 V/m, which comes out to be 6.03 x 10^-4 m squared. Since the capacitor has circular plates, I found the radius by taking a square root of A/pi, and it came out to be 0.014 m.

Does it look correct? I don't feel confident about that whole extrapolation...but then again, do I ever feel confident about physics?

And here's the next one:

2) The coil in a loudspeaker has an inductance of L = 112 microH. To produce a sound of frequency 40kHz, the current must oscillate between peak values of + and - 4.4 A in half a period. What average back emf is induced in the coil during this variation?

My main confusion is, why do we need to know the half period thing...I have no idea how to incorporate it into the solution.

Here's what I have now:

Xl = inductive reactance = 2 pi f L = 2 pi 40 x 10^3 x 112 x 10^-6 = 28.15 Ohm.

Then I rms = I/sqrt 2 = 4.4/sqrt 2 = 3.11 A. Am I correct in looking at rms values here?

Then V rms = I rms x Xl = 3.11 x 28.15 = 87.5 V.

Is that all? Is this V rms the average induced back emf they are asking for? Or am I wrong? I'm really confused about this one.

Thanks!

 

[jochi1543]

Nm

 

[mpatricksweeney]

hello,
with regard to pulleys...
Blocks A and B are connected across a frictionless pulley by a weightless cord. Block B lies on a horizontal frictionless surface and block A hangs vertically. What can be said about the acceleration of block B?

My confusion:
If B is resting on a frictionless surface, how does it create tension in the rope to resist the downward force exerted by A? When we draw a FBD, the net vertical force on B is 0 (since weight and normal cancel). Since the surface is frictionless, there is no horizontal force resisting the tension created by A (due to gravity). T is exerted on B in the horizontal direction, so in order for B to resist T an opposing horizontal force would need to exist. It's counterintuitive, but it seems like-- on an ideal frictionless surface-- whether B has the mass of a piano or a pencil, both will accelerate across the table at the same rate, as neither has "traction" to resist (to any degree) the downward force exerted by gravity on A. In which case, both B and A would accelerate downward at the rate, g.

This is not correct, but I can't account for the force which restrains B on the horizontal surface. The correct answer is: B would accelerate at some constant not equal to 9.8m/S^2.

Thanks for clarifying...

 

[gridiron]

hello,
with regard to pulleys...
Blocks A and B are connected across a frictionless pulley by a weightless cord. Block B lies on a horizontal frictionless surface and block A hangs vertically. What can be said about the acceleration of block B?

My confusion:
If B is resting on a frictionless surface, how does it create tension in the rope to resist the downward force exerted by A? When we draw a FBD, the net vertical force on B is 0 (since weight and normal cancel). Since the surface is frictionless, there is no horizontal force resisting the tension created by A (due to gravity). T is exerted on B in the horizontal direction, so in order for B to resist T an opposing horizontal force would need to exist. It's counterintuitive, but it seems like-- on an ideal frictionless surface-- whether B has the mass of a piano or a pencil, both will accelerate across the table at the same rate, as neither has "traction" to resist (to any degree) the downward force exerted by gravity on A. In which case, both B and A would accelerate downward at the rate, g.

This is not correct, but I can't account for the force which restrains B on the horizontal surface. The correct answer is: B would accelerate at some constant not equal to 9.8m/S^2.

Thanks for clarifying...

Hey! The first thing you need to do is to identify the system. When you say that weight and the normal force "cancel" you are correct but incorrect with regards to the system at hand. The system comprises the massless pulley, weightless cord and blocks A and B. This is because the analysis is from the perspective of acceleration. The normal force does not contribute to acceleration of block B. This is important for the MCAT because when you encounter a problem it is nice to draw all the forces for a free body diagram but most of the time you will be dealing with forces which contribute to the acceleration of an object either vertically or horizontally.

If I understand your question correctly, then block B can only move horizontally. This means that you can ignore the forces acting vertically on the block--this is the weight of the block and the normal force because those forces do not contribute to the acceleration of the block. You ask how there can be tension created by block B on the rope? In order for this system to be at rest or for either block to move, block B has to give some "pull" on the rope. This depends on the masses of the blocks. If block A is the same mass as block B, the system is at rest but there is still some tension in the rope created by block A and B to keep the system at rest. It is like a tug o war where both block have produce some "pull" on the rope. Since both blocks are the same mass, nothing moves because they have the same "pull" on the rope. What if block A was more massive? Then block B would slide across the surface. This is because block A has a greater "pull" on the rope resulting in block A accelerating in the direction of the net force. Block B still creates tension in the rope. Remember Newton's first law--an object at rest will stay at rest unless acted upon by some external force. Block B wants to stay on the surface so it produces a "pull" on the rope, but like tug o war, block A has a greater "pull."

To understand why the acceleration isn't g, you can use Newton's second law. For pulley problems on the MCAT, you should make an assumption as to which direction the system is going to accelerate if you want to find the acceleration. This can be tedious process but the MCAT is multiple choice. Instead of deriving the equation, for each choice plug in a value where both masses are equal and where you get zero you have your answer.

I hope I could be of some help. Both objects have to give some "pull" in order for the system to be at rest or for the system to accelerate. You have to account for forces which are involved in the motion of the object. I hope this helps and good , but post if you want more clarification.

 

[mpatricksweeney]

thanks a lot, grid. you are "being the change;" thanks for being so helpful.

I get the tug of war analogy you used, and certainly have observed this A/B system in nature, EXCEPT never on a frictionless surface. That's where I'm still hung up. I can't understand how B (on horizontal surface) participates in the tug of war without traction or exerting its own resistant force.

i guess the answer is simply that the downward Fg - T = mA*a exerted by A due to gravity accelerates B on the table surface; and the force generated by A (mA*g)must be opposed by a force of equal magnitude exerted by B (due to Newton's second, I think). In which case, if mB >>> mA, B will not accelerate. Is that reasonable?

Thanks again.

 

[gridiron]

thanks a lot, grid. you are "being the change;" thanks for being so helpful.

I get the tug of war analogy you used, and certainly have observed this A/B system in nature, EXCEPT never on a frictionless surface. That's where I'm still hung up. I can't understand how B (on horizontal surface) participates in the tug of war without traction or exerting its own resistant force.

i guess the answer is simply that the downward Fg - T = mA*a exerted by A due to gravity accelerates B on the table surface; and the force generated by A (mA*g)must be opposed by a force of equal magnitude exerted by B (due to Newton's second, I think). In which case, if mB >>> mA, B will not accelerate. Is that reasonable?

Thanks again.

Look at response below

 

[jochi1543]

Anyone have any clues about my problems? I took another jab at the third one today after I rested a bit, and lo and behold I solved it (it was pretty easy, as I suspected). I'm still in search of advice on the 1st and 2nd problems (in my first post at the top of this page).

 

[Foghorn]

Anyone have any clues about my problems? I took another jab at the third one today after I rested a bit, and lo and behold I solved it (it was pretty easy, as I suspected). I'm still in search of advice on the 1st and 2nd problems (in my first post at the top of this page).

I think the 1st problem is more complicated. Since it's charging, the current I isn't constant and you have to use I = delta Q divided by delta t, to give I dt = delta Q. Since you're given the time interval, you can find the total charge of the plates at 1 sec.....that's how I would approach it.

I don't know, I forgot all my physics already.

 

[tennisboy85]

The first case is when the system would be at rest.

How can B stay at rest if it's attached to a rope that's pulling it while it's on a frictionless surface? The only way it would stay "at rest" if there's static friction that's holding it in place, isn't it? Because we're talking about a perfectly flat table here.

 

[tennisboy85]

Another question: Can someone explain molecular orbitals to me? I did a practice test where I bombed every single question on a passage on this topic, I guess I don't understand it

 

[gridiron]

How can B stay at rest if it's attached to a rope that's pulling it while it's on a frictionless surface? The only way it would stay "at rest" if there's static friction that's holding it in place, isn't it? Because we're talking about a perfectly flat table here.

B wil stay at rest if there is nothing attached to it--that is what I meant by that comment.

mpatrick--the surface that block B is on is frictionless. If the block were to be pushed slightly, it will begin to accelerate in the direction the force was applied. No matter the mass of block A, block B will accelerate because there is no static or kinetic friction present. As an example, consider a ice puck on a frictionless surface. If you were to apply a push force, the puck will move and won't stop until an external force acts on it. This is why block B will accelerate. The reason there is tension in the rope due to B is that according to Newton's first law, the block wants to stay still. I hope this helps and good .

 

[RedSox1982]

This question is not out of a book or anything, but more of a conceptual problem I'm having. It regards elastic potential energy and gravitational potential energy. Here we go. I'll refer to "delta x" as dx.

Imgaine you have a 2kg mass on a weightless spring (k=.5). Originally the mass is held by a board at the spring's point of origin (dx = 0), and then the board is removed.

With this, you can find dx:
mg = k(dx)
So dx = 40m. Great.

Now what I want to know is this... the change in GRAVITATIONAL potential energy (mgh) is 800J. One can assume that the grav. potential energy was converted to kinetic energy and then to elastic potential energy, such that that same 800J should now be completely converted to elastic potential energy (assuming no loss due to non-conservative forces). BUT! When I use the elastic potential energy equation...

(1/2)k(dx)^2 = 400J.... NOT 800J! How can this be?!!? These have to be the same, don't they?! The other 400J didn't just disappear... Looking at it again, you see that you assumed...

mg(delta h) = (1/2)k(dx)^2
Since h = dx, we can see that
mg = (1/2)k(dx)

But mg = k(dx)... that's Hooke's Law! Can someone help me out here?? Where did I go wrong? What have I missed?

 

[ashley]

I have the following rule from class notes:

wave speed is a property of the medium and does not depend on the frequency.

is it not possible, then, to increase wave speed by increasing frequency? (according to v=wavelength*frequency)

 

[gridiron]

I have the following rule from class notes:

wave speed is a property of the medium and does not depend on the frequency.

is it not possible, then, to increase wave speed by increasing frequency? (according to v=wavelength*frequency)

Hey! A wave is a disturbance which moves through a medium. The disturbance moves through the medium. What is important in understanding waves is the medium. The medium can be anything--it can be air, water, rock, car...etc. What differentiates each medium is the physical properties of the medium--the make-up of each medium. Each medium has a different density, elasticity, and tension and the disturbance must propogate through each. What differentiates each disturbance, wave, through each medium is the amplitude, frequency and wavelength--these are intrinsic properties of the wave. The speed of a wave is determined by how much area a wave covers over a period of time. Depending on the properties of the medium, waves will cover different distances over different times. This is why the speed of the wave is dependent on the medium in which it travels and not on the intrinisic properties of the wave--the amplitude, frequency and wavelength. Changing the frequency of the wave only changes the wavelength of the wave. You do use both the frequency and wavelength to calculate wave speed, but that is only for a given medium. I hope this helps and good .

 

[gridiron]

This question is not out of a book or anything, but more of a conceptual problem I'm having. It regards elastic potential energy and gravitational potential energy. Here we go. I'll refer to "delta x" as dx.

Imgaine you have a 2kg mass on a weightless spring (k=.5). Originally the mass is held by a board at the spring's point of origin (dx = 0), and then the board is removed.

With this, you can find dx:
mg = k(dx)
So dx = 40m. Great.

Now what I want to know is this... the change in GRAVITATIONAL potential energy (mgh) is 800J. One can assume that the grav. potential energy was converted to kinetic energy and then to elastic potential energy, such that that same 800J should now be completely converted to elastic potential energy (assuming no loss due to non-conservative forces). BUT! When I use the elastic potential energy equation...

(1/2)k(dx)^2 = 400J.... NOT 800J! How can this be?!!? These have to be the same, don't they?! The other 400J didn't just disappear... Looking at it again, you see that you assumed...

mg(delta h) = (1/2)k(dx)^2
Since h = dx, we can see that
mg = (1/2)k(dx)

But mg = k(dx)... that's Hooke's Law! Can someone help me out here?? Where did I go wrong? What have I missed?

Hey! I think there is a error in the type of potential energy you first calculate. Potential energy is the energy of an object due to its position to some reference--an object will gain potential energy when it moves from its reference against a opposing force. If the board is suddenly removed, then the mass will fall on the spring. The spring will gain energy in the form of elastic potential energy. This is because the spring is displaced by the weight of the mass from its reference point against the restoring force, given by hooke's law, acting on the spring. After the spring is fully compressed, by the distance you calculated, the elastic potential energy will be converted into gravitational potential energy when the mass is shot into the air---you can calculate the height by setting the gravitational potential energy equal to the elastic potential energy because energy will be converted between the two. When the spring is compressed, the system of the mass and the spring cannot possess both gravitational and elastic potential energy. I hope this helps and good .

 

[RedSox1982]

I COMPLETELY understand what you are saying regarding the conservation of energy... but the numbers don't check out. In my problem, the dx=40. That's going to be the same HEIGHT that is diminished - the potential energy lost will be mgh, and the kinetic potential energy gained will be 1/2k(dx)^2.

Setting these two things equal, you get:
mg(dh)=1/2k(dx)^2, where dh and dx are the change in height and the change in distance of compression of the spring, which are the same, so...
mg=1/2k(dx) - because dx = dh
BUT!
Hooke's Law states that mg = k(dx)... right? Didn't I just show that NOT to be true? I still don't see where my logic is wrong.

 

[mpatricksweeney]

4microC of negative charge are transferred from one plate of an 8 microF capacitor to the other plate. How much work was done by the electric field during the charging process?

a -4 microJ
b -1 (answer)
c 1
d 4
E-field---->
emf---------+| |(-)--------------
<-----e-

The magnitude is fine (dPE=W=.5CV^2), but I'm unsure why the work done is negative. Bear w me, please: the negative charges are moving from the negative plate toward the positive plate against the E-field (which is + to -), correct? In that case, they're moving down their electric potential (ie toward positive plate). As a result of moving down their potential, transferred electrons (now on the positive plate) have less potential energy. The potential energy they've lost is represented by the negative sign (exergonic)? Is this a correct understanding?

What kind of force/agent would transfer the e-? What triggers the transfer? The book says the negative work is done by the electric field-- as in qEd? Total circuit energy is less following the transfer; what would be required to restore transferred e- to the neg plate?

Too abstract...thanks a lot.

 

[tennisboy85]

Question: Examkraker says that intensity of light wave is proportional to the square of the amplitude. Can someone explain why?

Thanks a bunch!

 

[tennisboy85]

Another Q:

On a hot day, mirage may appear as puddle of water far ahead on hot pavement. The mirage is actually a virtual image of the sky caused by light rays bending. Which of the following helps to explain the formation of the mirage?

Answer: As air warms the speed of light increases.

I don't quite understand the logic behind this, and why would speed of light increase in warm air?

Thanks for the help.

 

[tennisboy85]

Hi Redsox, I thought about your question and you can't include the springs into this picture assuming that the distance stretched will be 20 m up from equilibrium and 20 m down from equilibrium.

Your example is like a bungee jumper, so when the person jumps - there's only potential energy (mgh)

at mid point, there's mgh + kinetic energy

at the bottom, there's only spring elastic energy because the potential energy and kinetic energy have been absorbed by the spring.

What you need to understand is: the X from equilibrium midpoint to the bottom (where v = 0) would be longer than X from equilibrium midpoint to the top (where the jumper first started).

Therefore, spring elastic energy from mid point to bottom is much greater than spring elastic energy from the top to the midpoint (which explains for the "missing energy" you were mentioning)

 

[mpatricksweeney]

Temperature of an ideal gas undergoing free adiabatic expansion should:
a remain constant
b increase b/c the molecules will move faster as they escape into the vacuum
c decrease b/c the molecular energy is spread over a larger area
d decrease because the pressure goes down

I selected c, although the rationale is lousy. The answer in the book is A, the temperature will remain constant because "no work is done and no heat escapes or enters the system." If we have adiabatic process, Q=0J. If Q=0, and we have an expansion, positive work must be done by the gas, right? In which case, the internal energy of the system decreases (ie U = Q-W), so we expect the avg KE of the molecules to decrease, and the temperature therefore.

Does adiabatic always only mean no heat transfer (Q=0), or does adiabatic tell us something about the work done?

Thanks a lot.

 

[smeagol]

This question is not out of a book or anything, but more of a conceptual problem I'm having. It regards elastic potential energy and gravitational potential energy. Here we go. I'll refer to "delta x" as dx.

Imgaine you have a 2kg mass on a weightless spring (k=.5). Originally the mass is held by a board at the spring's point of origin (dx = 0), and then the board is removed.

With this, you can find dx:
mg = k(dx)
So dx = 40m. Great.

Now what I want to know is this... the change in GRAVITATIONAL potential energy (mgh) is 800J. One can assume that the grav. potential energy was converted to kinetic energy and then to elastic potential energy, such that that same 800J should now be completely converted to elastic potential energy (assuming no loss due to non-conservative forces). BUT! When I use the elastic potential energy equation...

(1/2)k(dx)^2 = 400J.... NOT 800J! How can this be?!!? These have to be the same, don't they?! The other 400J didn't just disappear... Looking at it again, you see that you assumed...

mg(delta h) = (1/2)k(dx)^2
Since h = dx, we can see that
mg = (1/2)k(dx)

But mg = k(dx)... that's Hooke's Law! Can someone help me out here?? Where did I go wrong? What have I missed?

I think I see what is bothering you.

It seems that you're to confirm the derivation of Hooke's law via some energy equation. That is:

mgx = 0.5kx^2

Divide by x:

mg = .5kx

mg is Force, but you don't get Hooke's law from this (Hooke's law: F = -kx), and that seems to be what is upsetting you. I can't explain without using some calculus, but it's not too bad:

Hooke's law is the DERIVATIVE of the energy equation for elastic potential energy. So to get hook's law try differentiating that equation.

Take the derivative of both sides:

W = 0.5kx^2

-differentiate (instead of dividing by x). Recall that mg is a constant.

F = kx

-Now that's Hooke's law!

So when you springs and energy, don't use Hooke's law (which gives you force) unless you plan to integrate the force to get work done (energy in Joules).

If you're interested, here's how I would have approached the problem:

Using conservation of energy: Initial energy = Final energy


Initial potential energy from gravitation: mgx
Initial kinetic energy = 0
Initial elastic potential energy: 0 (not compressed yet)

Final potential energy from gravitation: 0 (when it's done compressing, I set the height = 0)
Final kinetic energy: 0
Final elastic potential energy: .5kx^2

So if initial energy = final energy, then:

mgx = .5kx^2

2mg/k = x; x = 80m; Given that m = 2, k = 0.5, g = 10, all in SI units.

edit: Removed conclusion mg = kx (not sure if this move is valid)

mg is a constant force and it's not fair to set it as the F in a Hooke's law equation. Imagine if you set x = 10000000.

mg will still be = mg, however, kx generates a force that varies by the length of compression, it will enlarge.

So in a Hooke's law equation, you can't let mg (a constant force) = kx (varying force)

 

[ouchitburns]

ignore

 

[axp107]

How is velocity or sound or any wave affected by density..

I thought, an increase in density means an increased speed.. and that sound travels faster thru solid than air or water...

however, EK says speed decreases as density increases

 

[mpatricksweeney]

How is velocity or sound or any wave affected by density..

I thought, an increase in density means an increased speed.. and that sound travels faster thru solid than air or water...

however, EK says speed decreases as density increases

Inverse relationship:
velocity = sqrt [(4/3(Shear mod) + (Bulk mod) ] / density ]

Density's in denominator. As density increases, velocity decreases.

There's the formula, but can anyone explain why it's so?

 

[axp107]

I always thought sound travels faster in solids than in water or air.

 

[ouchitburns]

Hey,

does friction do work? I have heard from a number of sources that it does not, but then I got this answer from the AAMC test 4:

Work is defined as the product of external forces exerted on an object times the distance the object moves parallel to the forces. In this question, the forces on the block are the string tension and the kinetic friction between the block and board; the former is parallel to the movement of the block on the board, and the latter is antiparallel.

Also, does gravity do negative work?

 

[byeh2004]

Can your level arm be your position vector? Can it be the same thing? When is it the same thing? Thanks!

 

[smeagol]

Hey,

does friction do work? I have heard from a number of sources that it does not, but then I got this answer from the AAMC test 4:

Also, does gravity do negative work?

1) I believe friction does indeed do work. The unit of work is the joule or a Newton meter. I haven't done the AAMC's yet so I didn't read what they wrote.

When an object slides across a frictional surface, there is a force of friction, f, acting upon it (in Newtons). For it to do work, it needs to slide a certain distance, d. So imagine the force of friction, f over some distance, d. The result is a Newton-meter, or a Joule in terms in energy. Hence, there is some work done.

2) From what I understand, gravity can do both negative work AND positive work, depending on the direction the object of interest is moving.

Recall the formula: Fdcos(theta).

Considering the force of gravity, it can be rewritten as: mgcos(theta)

The part of the formula that determines whether work is positive or negative depends on cos(theta), and theta is the angle between the force vector acting on the object and the direction in which it is moving.

Hence, if an object begins at some height, h, and falls down, it moves in the SAME direction as the force of gravity. Therefore, the angle between the d and F is 0 degrees. So cos(0) = 1. Work is positive.

Gravity will do negative work if the object is moving in the opposite direction of the gravitational force. So if the object is moving 'up' while gravity is pulling it 'down,' theta will be 180 degrees. Since cos(180) is -1, work done by gravity will be negative. *Note that there can be other forces acting on the object that may be doing positive work.

-------------

@Ouchitburns: I have a question for you

You posted a question with springs that baffled me but you edited it. I was quite baffled by the question. If you can, please, PM me, or repost the question/explanation you found for it because it's been bugging me.

I think it was the one with 2 forces acting on both sides of a spring that compresses.

-Thanks.

 

[smeagol]

Can your level arm be your position vector? Can it be the same thing? When is it the same thing? Thanks!

This might not be too clear without a diagram but yes, the lever arm can indeed sometimes be equal to the position vector. I think that this the case when your system of interest currently a straight line.

The lever arm, d, is given by the equation d = rsin(theta), where r is the position vector and theta is the angle between r and the applied force, F. When the position vector, r, is orthogonal to the applied force, sin(90) will of course be 1.

So d = r when the applied force is perpendicular to the positive vector, r.

 

[dill_punjabi]

Can any one explain me Examcracker 577?

 

[gridiron]

Can any one explain me Examcracker 577?

Hey! I, and many others, don't have the examcracker books. Could you please post the question in its entirety? Thanks.

 

[axp107]

Can anyone explain if sound travels faster through water or solid?

 

[MundaneMD]

Can anyone explain if sound travels faster through water or solid?

Sound travels faster in solids than in liquids.
Sound waves are longitudinal, so they travel faster when a material is more dense (i.e. solids).

 

[axp107]

How does friction work in "rolling" and "translational" motion.. how are they different?

Is it an important distinction to make for the MCAT?

 

[thechairman]

Hi, I have a question on electro-statics. I know that two like charges will repel each other due to their electric field.

Say there is a hypothetical situation where there are two particles of like charge, one is stationary, and the other is not. So my question is, since both particles feel the repulsion force given by coulomb's law, the non-stationary charge will accelerate away from the stationary charge (due to the force), but does its velocity ever become zero? Since the only acceleration is from the repulsive force, does the moving particle continue to move away from the stationary charge until their distance is infinity?

But if you look at the problem from another angle, that of the conservation of energy. Set E*q*d = 1/2 * m * v^2, and from this equation we know that the particle's electrical potential energy is being converted to kinetic energy, as it moves away from the stationary particle. But since the electrical potential is finite, we know that energy is eventually exhausted and the particle must come to a stop.

So which perspective is right? They seem to contradict eachother, or am I missing something here?








My contribution to thread:

How does friction work in "rolling" and "translational" motion.. how are they different?

Is it an important distinction to make for the MCAT?

From what I know, in rolling, frictional force is in the direction of motion. Since friction is defined as the resistance to sliding, friction must act forward (since the wheel is turning backwards to cause the car to move forward) to prevent the wheels from sliding.

 

[jochi1543]

This is simple, and I'm pretty sure it's a mistake on my prof's part, actually, but I wanted to run this by SDNers. It's a multiple choice problem from the sample physics midterm (my real midterm is on Monday). There are 2 wires perpendicular to each, carrying the same current in the direction as shown. Show at which point - a, b, c, d, or e - the net magnetic field is zero.

At points a, b, and d both fields point in the same direction, so nothing cancels out. At c and e, however, they point in opposite directions (B due to wire 1 is into the page and B due to wire 2 is out of the page), so it yields a net magnetic field of zero for both points. However....there's only one reply! The reply is c. Where did I make a mistake with e? C and e are on the same side of the two wires, so how could the net field at c be zero, but not the net field at e?

It's too late to find out the answer from the prof now, but hopefully someone here can either confirm that it's his mistake or correct my interpretation.

 

[MundaneMD]

Points C and E are different distances away from Wire 2.

It looks like for Point C, the distance from Wire 1 = the distance from Wire 2. So the net field would be 0.

Point E looks to be about twice the distance away from Wire 2 than Wire 1. The magnetic field gets weaker depending on distance (don't remember the exact formula). So even though the two magnetic fields are in opposite directions, they are not equal. Thus, the net field is not 0.

 

[MundaneMD]

Hi, I have a question on electro-statics. I know that two like charges will repel each other due to their electric field.

Say there is a hypothetical situation where there are two particles of like charge, one is stationary, and the other is not. So my question is, since both particles feel the repulsion force given by coulomb's law, the non-stationary charge will accelerate away from the stationary charge (due to the force), but does its velocity ever become zero? Since the only acceleration is from the repulsive force, does the moving particle continue to move away from the stationary charge until their distance is infinity?

But if you look at the problem from another angle, that of the conservation of energy. Set E*q*d = 1/2 * m * v^2, and from this equation we know that the particle's electrical potential energy is being converted to kinetic energy, as it moves away from the stationary particle. But since the electrical potential is finite, we know that energy is eventually exhausted and the particle must come to a stop.

So which perspective is right? They seem to contradict eachother, or am I missing something here?

An object in motion tends to stay in motion. Unless there is another force stopping it, the particle will keep on moving away.

The moving particle will initially gain a lot of acceleration because its close to the stationary particle. As it gets further away, it's acceleration will approach 0 m/s². But, its velocity will keep it moving. Until infinity.

 

[jochi1543]

Points C and E are different distances away from Wire 2.

It looks like for Point C, the distance from Wire 1 = the distance from Wire 2. So the net field would be 0.

Point E looks to be about twice the distance away from Wire 2 than Wire 1. The magnetic field gets weaker depending on distance (don't remember the exact formula). So even though the two magnetic fields are in opposite directions, they are not equal. Thus, the net field is not 0.

Ahhhh! Makes sense.

 

[axp107]

Electric power for long distance transmission is stepped up to a very high voltage to:

a) produce currents of higher density
b) produce higher currents in transmission wires
c) to make less insulation necessary
d) to cut down the heat loss in transmission wires


Answer is D).. can anyone explain?

 

[thechairman]

since P = V*I, and the power loss from resistance is P = I^2 * R, you want to minimize current and maximize the voltage, in order to reduce the power loss.

 

[axp107]

Why is power loss from resistance P = I^2 * R

P = IV and P = 1^2 * R is the same thing.. just written with different variables right?

 

[eltor07]

I don't understand how increasing Voltage decreases current AND resistance. If P = iV then yes with P being constant inc V will dec i but then if u look at the relationship b/w resistance and current with P = i^2R you see that decreasing current would increase resistance.

What's the conceptual aspect that makes this all work?

Edit: Unless resistance is also constant! Then that makes sense cuz then decreasing current would mean that Ploss would be lower due to the resistor which really happens because current is decreased. That makes sense right? Because resistance is intrinsic to the material used not whether there is more current.

 

[eltor07]

That would mean that by increasing voltage, less current is sent through, but also less is loss because of that; instead of having lots of current, but an increased amount of loss?