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]

Is the answer 200?

Let's see what we know so far...

V initial
V final = 0

thats it! we know we somehow have to use the kinematics equations.. b/c we need distance right?

Let's look at V^2 = V0^2 + 2ad

We don't have acceleration though. BUT we have information about the frictional force...

At first glance, it seems like we can't use the formula:

Friction force = Normal Force * Coefficient of Friction

But we can! Let's see what happens! What's the ONLY force acting on the car in this problem? FRICTION only! Nothing else... SO we can make:

Friction Force = ma

So, ma = mg * 0.1

The M's cancel out, giving an acceleration of 1 m/s^2

Plug acceleration in: V^2 = V0^2 + 2ad
and there's your answer

 

[jochi1543]

EDIT: nevermind, I think I got it

 

[axp107]

One thing that I never knew the answer to and always bugged me:


What if an object is moving at a constant velocity, would STATIC or KINETIC friction act on it?

 

[gridiron]

One thing that I never knew the answer to and always bugged me:


What if an object is moving at a constant velocity, would STATIC or KINETIC friction act on it?

Hey! Constant velocity means the acceleration is zero. By Newton's second law this would mean the net force acting on the object is zero. Since static or kinetic friction are forces which act to impede an objects motion, if the net force acting on a object is zero, static or kinetic friction would not act on the object. If they did, the velocity wouldn't be constant. Hope this helps and good .

 

[axp107]

Hey! Constant velocity means the acceleration is zero. By Newton's second law this would mean the net force acting on the object is zero. Since static or kinetic friction are forces which act to impede an objects motion, if the net force acting on a object is zero, static or kinetic friction would not act on the object. If they did, the velocity wouldn't be constant. Hope this helps and good .

This question was asked earlier.. according to the question, friction was still acting on the car even though it was moving as constant velocity:

When a car moves under its own power at constant velocity, the frictional force between the road and the tires that propel the car is:

A) static and in the direction of the motion of the car - CORRECT

 

[jochi1543]

Hey! Constant velocity means the acceleration is zero. By Newton's second law this would mean the net force acting on the object is zero. Since static or kinetic friction are forces which act to impede an objects motion, if the net force acting on a object is zero, static or kinetic friction would not act on the object. If they did, the velocity wouldn't be constant. Hope this helps and good .

However....when a car is moving at constant velocity (think cruise control), there IS friction acting on it - if you turn off the engine, it will come to a stop. Or is this somehow not an applicable example?

 

[axp107]

However....when a car is moving at constant velocity (think cruise control), there IS friction acting on it - if you turn off the engine, it will come to a stop. Or is this somehow not an applicable example?

I think gridiron might have misunderstood my question...

Yeah like the problem I mentioned on the other page... a car moving at constant velocity for example, friction has to be there between the tires right?

Is this friction static friction

or

is it kinetic?

Does kinetic apply/not apply when acceleration is not happening?

 

[MundaneMD]

Note: I might be wrong on this, so correct me if I'm wrong.

However....when a car is moving at constant velocity (think cruise control), there IS friction acting on it - if you turn off the engine, it will come to a stop. Or is this somehow not an applicable example?

Yes, you're correct. When you turn on cruise control, the car is accelerating. But, since there is kinetic friction, which opposes any type of motion, the two forces cancel out. You're left with a car moving with no net force in that direction, which is considered to be constant velocity.

Yeah like the problem I mentioned on the other page... a car moving at constant velocity for example, friction has to be there between the tires right?

Is this friction static friction

or

is it kinetic?

Does kinetic apply/not apply when acceleration is not happening?

Static friction: the force that must be overcome to set an object in motion.
Kinetic friction: opposes the motion of objects moving relative to each other.

Having a constant velocity means that the car is in motion. The car has to accelerate (step on the gas) to cancel out the force of kinetic friction. This is why, if you take your foot off the gas, your car slows down. Kinetic friction exists when you rub two items together (like tires on the road). If you don't believe me, touch your tires the next time you finish driving. Or better yet, imagine driving on a surface where the wheels weren't opposed by kinetic friction (e.g. ice). Good luck trying to stop your car.

 

[jochi1543]

Done

 

[jochi1543]

Note: I might be wrong on this, so correct me if I'm wrong.



Yes, you're correct. When you turn on cruise control, the car is accelerating. But, since there is kinetic friction, which opposes any type of motion, the two forces cancel out. You're left with a car moving with no net force in that direction, which is considered to be constant velocity.



Static friction: the force that must be overcome to set an object in motion.
Kinetic friction: opposes the motion of objects moving relative to each other.

Having a constant velocity means that the car is in motion. The car has to accelerate (step on the gas) to cancel out the force of kinetic friction. This is why, if you take your foot off the gas, your car slows down. Kinetic friction exists when you rub two items together (like tires on the road). If you don't believe me, touch your tires the next time you finish driving. Or better yet, imagine driving on a surface where the wheels weren't opposed by kinetic friction (e.g. ice). Good luck trying to stop your car.

Actually, when the car is moving at a constant speed, the friction between the tire and the road is static. Don't forget, the tire is rotating, so instead of sliding along the pavement, a new section of the tire comes into contact with the road at every moment.

However, if you hit the brake, your wheels lock and stop rotating, THEN you have kinetic friction - the same spot on the tire is now sliding against the pavement.

 

[axp107]

Actually, when the car is moving at a constant speed, the friction between the tire and the road is static. Don't forget, the tire is rotating, so instead of sliding along the pavement, a new section of the tire comes into contact with the road at every moment.

However, if you hit the brake, your wheels lock and stop rotating, THEN you have kinetic friction - the same spot on the tire is now sliding against the pavement.

so:

static friction = when not moving or no acceleration

kinetic = when there is acceleration?

 

[jochi1543]

so:

static friction = when not moving or no acceleration

kinetic = when there is acceleration?

Yeah, no acceleration between the 2 surfaces in question.


Also, no acceleration and subsequently zero net force doesn't mean there's no friction - it just means that the force of friction (if present) is exactly equal to whatever force propels the object forward (if such force is present).

 

[MundaneMD]

I give up. I'll stay away from friction problems from now on.

 

[BlackSails]

Actually, when the car is moving at a constant speed, the friction between the tire and the road is static. Don't forget, the tire is rotating, so instead of sliding along the pavement, a new section of the tire comes into contact with the road at every moment.

However, if you hit the brake, your wheels lock and stop rotating, THEN you have kinetic friction - the same spot on the tire is now sliding against the pavement.

That is not correct. There is no acceleration between the two surfaces, but there is definatly movement between the two. Moving in relation to the surface = kinetic.

In this case, since the wheels roll, you use rolling friction.

 

[axp107]

Hey guys,

Anyone have a link to how contact lenses work in terms of optics?

I know for nearsightedness you need a diverging lens to correct the problem (image in front of retina)... and for farsightedness (image behind retina), you need a converging lens to correct the problem.

For example, diverging lenses form an image on the same side as the source of light .. always.. how would this help with nearsightedness?

I'm having trouble picturing how the lenses would help.. thanks!

 

[chad5871]

In a completely inelastic collision (i.e. the two objects stick together), what happens to the lost kinetic energy? This website seems to imply that it's lost mostly as heat and sound. Is that correct?

From the site: "The missing energy is transformed mainly into heat and sound, so the total kinetic energy of the system after contact is less than it was before."

 

[BlackSails]

In a completely inelastic collision (i.e. the two objects stick together), what happens to the lost kinetic energy? This website seems to imply that it's lost mostly as heat and sound. Is that correct?

From the site: "The missing energy is transformed mainly into heat and sound, so the total kinetic energy of the system after contact is less than it was before."

Heat, sound, deformation of the objects, vibrational modes, etc. Its just no longer kinetic energy.

 

[chad5871]

Heat, sound, deformation of the objects, vibrational modes, etc. Its just no longer kinetic energy.

Sweet. Thanks so much for the response.

 

[HristosKaran]

Stupid Kaplan problem...

Which of the following is the least amount of information that could be used to determine how far a cannonball lands from a cannon?

A initial velocity, angle of elevation

B initial velocity, time to travel, angle of elevation

C initial velocity, acceleration of gravity

D initial velocity, angle of elevation, acceleration of gravity

E initial velocity, height of the peak of the projectile

Answer is C...kaplan says x=vot+1/2at2 was used...but how did they figure out the time from this problem? Did the just assume they would know the time??

 

[BlackSails]

Stupid Kaplan problem...

Which of the following is the least amount of information that could be used to determine how far a cannonball lands from a cannon?

A initial velocity, angle of elevation

B initial velocity, time to travel, angle of elevation

C initial velocity, acceleration of gravity

D initial velocity, angle of elevation, acceleration of gravity

E initial velocity, height of the peak of the projectile

Answer is C...kaplan says x=vot+1/2at2 was used...but how did they figure out the time from this problem? Did the just assume they would know the time??

If you assume that the cannon and the target range are on level ground, t can be easily calculated from the velocity and g. Simply find the time of flight of the cannon ball, and then you can find how far the ball goes in that time.

 

[estairella]

If you assume that the cannon and the target range are on level ground, t can be easily calculated from the velocity and g. Simply find the time of flight of the cannon ball, and then you can find how far the ball goes in that time.

Umm... what? Unless you're assuming you know the angle of elevation of the initial velocity, you won't be able to get the flight time. It's just a crappy question overall, the minimum information you need to solve the problem is:
a) initial velocity (magnitude & direction) and gravitational acceleration OR initial horizontal velocity and flight time OR initial velocity (magnitude) and height of peak and gravitational acceleration
b) relative heights of the cannon and landing ground (not needed if you know horizontal velocity & flight time)
c) drag due to air resistance

Now, because this is the MCAT, you should be able to assume, unless otherwise stated:
- g = 9.8m/s^2
- level ground
- no air resistance

If you think "initial velocity" should mean both magnitude and direction, then really, initial velocity is all the information you need. If you take it to mean you also need angle of elevation information, then both A and E would be correct. Imo it's bad form to expect "gravitational acceleration" as answer, since this implies that it could be any number. What if it's in outer space and gravitational acceleration is 0? ******ed...

 

[chad5871]

It's just a crappy question overall

Agreed. I think it's just a bad question.

 

[BlackSails]

Umm... what?

I meant in addition to the intial velocity and g.

 

[HristosKaran]

Here is another crapper..lol

A uniform rod 2 m in length with a mass of 4 kg is held at an angle θ (between 0° and 90°) to a frictionless tabletop by a string pulling straight up 0.5 m from the upper end of the rod. What is the tension in the string?


13.3 N

40 N

26.7 N

0 N

depends on the value of ?


Answer is 26.7N....There is a diagram, but it doens't really do much for the problem..if someone could explain how they got to this answer in lamen's terms, that would be great!!

 

[BlackSails]

Here is another crapper..lol

A uniform rod 2 m in length with a mass of 4 kg is held at an angle θ (between 0° and 90°) to a frictionless tabletop by a string pulling straight up 0.5 m from the upper end of the rod. What is the tension in the string?


13.3 N

40 N

26.7 N

0 N

depends on the value of ?


Answer is 26.7N....There is a diagram, but it doens't really do much for the problem..if someone could explain how they got to this answer in lamen's terms, that would be great!!

If the rod were standing straight up, the tension would be 40 N. If it were on the table, the tension would be 0. Therefore, the tension at the angle θ is between the two. So its either 13 or 26. Is the diagram to scale? If so, you use F=mg*cosθ, and guess at theta.

 

[neo123]

A question in a book that I think is weird..

it shows a circuit with a resistor first.. then a capacitor and asks to draw a graph of Resistance VS. Capacitance...

what would happen? would resistance stay constant as capacitance increases... or would it decrease?

 

[gridiron]

Bump

 

[HristosKaran]

Okay, I don't have a specific question, but I'm trying to understand some trends with kinematics.


If two objects of equal mass and velocity are launched at different angles...lets say A = 60 degrees, while B = 30 degrees. Do they both go an equal distance in the x-direction, just a difference in time occurs? Basically I'm trying to figure out how angle and mass affect projectiles in x & y directions... thanks

 

[cheezer]

Okay, I don't have a specific question, but I'm trying to understand some trends with kinematics.


If two objects of equal mass and velocity are launched at different angles...lets say A = 60 degrees, while B = 30 degrees. Do they both go an equal distance in the x-direction, just a difference in time occurs? Basically I'm trying to figure out how angle and mass affect projectiles in x & y directions... thanks

okay i looked up the kinematics equations. i think the one on the bottom right should answer your question.

for the x component, you have two different velocities. plug in Vi cos 60, Vf = 0 and d. then plug in a seperate Vi cos 30, Vf=0, and the same d value. so yes, they'll get to the same point, one will just take longer

 

[axp107]

Circuit rules that have been bugging me... from class... I've never had to deal with circuits and resistors in 1 circuit..

would voltage across resistor remain constant as voltage across capacitor increases/decreases as it is charged/discharged?

 

[fileserver]

A question in a book that I think is weird..

it shows a circuit with a resistor first.. then a capacitor and asks to draw a graph of Resistance VS. Capacitance...

what would happen? would resistance stay constant as capacitance increases... or would it decrease?

I am going to take a stab at this, don't know if I am right. V=Cq=IR. If C increases, R increases.

 

[fileserver]

okay i looked up the kinematics equations. i think the one on the bottom right should answer your question.

for the x component, you have two different velocities. plug in Vi cos 60, Vf = 0 and d. then plug in a seperate Vi cos 30, Vf=0, and the same d value. so yes, they'll get to the same point, one will just take longer

If you are talking about vi cos 60, then you are talking about the x component. In a projectile, the Vf is the same as Vi. I think you meant vf of the y direction and using Voy to calculate the time using the bottom left equation. Then using the bottom right equation, you would get the same d.

You're right. They both reach the same point, but one takes longer.

 

[bigserve99]

Quick Question from "The Nova - MCAT Physics Book"

Page 76 - Chapter 5 - Passage 1 (The only passage really) - Question 5:


So the passage basically is about centripetal acceleration and the force of gravity on a man standing on a scale at the equator or the poles of the planet. The relevent information is as follows from the passage:

"We have also assumed that the Earth is a perfect sphere. Because it is rotating, the distance from the center of the Earth to the equator is greater than the distance from center to pole by about 0.1%"

That's all great, but here's the question:

"If two identical men stood on scales at the south pole and at the equator of an Earth identical to this one but nonrotating, how would the reading on the polar scale compare to the equatorial one?

A. It would be less
B. It would be the same.
C. It would be greater.
D. There is not enough information to answer this question.

I answered B, they would be the same. My reasoning is that this Earth in the question is not rotating. The differential in radius between the pole and the equator exists because the planet is rotating, as per the passage. Without rotation, this differential should cease to exist. Therefore, I would contend the answer is B.

The answer given is C, citing that it has to with the radius differential. I mean, yah, that's obvious, but it doesn't take into account what was given in the passage, mainly that without rotation, I don't see how the differential exists in the first place.

Any thoughts?

Thanks.

 

[bigserve99]

Ok, so I just realized something.

The question does say "identical" to this "one"........so I suppose technically, this includes the radius differential between the the real earth and imaginary non-rotating earth to also be identical (though this makes no sense).

Man, they get really picky with this stuff.

 

[carolianamcat]

I have spent hours trying to figure out this problem and would be really grateful if someone could help me. The answer is written at the back but the ek book does not do a good job explaining how you can get the right answer.

question-
Person A drops a rock from a 100 m building. At exactly the same moment, person B throws a rock from the bottom of the building to the top of the building. At what height do the rocks meet?
(the answer is supposed to be 75m)

 

[prudvi]

I have spent hours trying to figure out this problem and would be really grateful if someone could help me. The answer is written at the back but the ek book does not do a good job explaining how you can get the right answer.

question-
Person A drops a rock from a 100 m building. At exactly the same moment, person B throws a rock from the bottom of the building to the top of the building. At what height do the rocks meet?
(the answer is supposed to be 75m)

did they give the velocity at which person B throws the ball. i don't know how to solve the problem with out knowing the initial velocity of person B

 

[carolianamcat]

Hey Prudvi,
Thanks for replying, I checked the book again and this is the exact question, no additional info such as velocity given.

 

[bigserve99]

Hey Prudvi,
Thanks for replying, I checked the book again and this is the exact question, no additional info such as velocity given.

Hi Carolinamcat,



- First of all, V0 for B is the same as Vf for A. The problem states that rock B is thrown to the same height as the top of the building. This means it is a standard projectile motion.

Based on this, and the fact that they were thrown at the same time0, we can immediately note the following:

The rocks must meet at the same

1. time
2. place (in 2space I mean)
3. with same velocity when they meet (though in opposite directions)

I chose to equate the final velocities when the meet, and by doing so, determine the time t at which these meet.

Vf = V0 + at

In addition, I chose to set up the problem such that negative acceleration was down (ie. g = -10 m/s^2)

For rock A (the one being dropped):
Vf = V0 + at becomes Vf = 0 - at which becomes Vf = -10t(notice Vf is negative)

For Rock B (the being thrown up)
Vf = V0 + at becomes Vf = V0 - at, but now we need V0.

well, in a projectile problem, the V0 of the upward trajectory is equal to the Vf of the downard portion of the problem, so V0 (of B) is equal to Vf (of A) which is nothing but sqrt (2gh) (from vf^2 = v0^2 + 2ax for free fall). I arrived at V0 = 20*sqrt(5) = approx. 44.7

so, for Rock B:
Vf = V0 - at becomes Vf = 44.7 - 10t
EQUATING THE TWO Vf's gives:

-10t = 44.7 - 10t

well, the -10t's seem to cancel, BUT, IN FACT THEY ARE POINTED THE OPPOSITE WAYS.....so to make them truly equal, you have to add a negative....hence,

10t = 44.7 - 10t which gives t = 44.7 / 20 = 2.235 seconds.

with t = 2.235 just plug and chug in to rock a or b ( I chose a):
x = v0t - 1/2 at^2 which gives -25 meters (with v0 = 0).

or just 25 meters from the top of the building down....or 75 meters.

Hope that made sense, carolinamcat.

 

[BerkReviewTeach]

Hi Carolinamcat,

- First of all, V0 for B is the same as Vf for A. The problem states that rock B is thrown to the same height as the top of the building. This means it is a standard projectile motion.

Based on this, and the fact that they were thrown at the same time0, we can immediately note the following:

The rocks must meet at the same

1. time
2. place (in 2space I mean)
3. with same velocity when they meet (though in opposite directions)

I chose to equate the final velocities when the meet, and by doing so, determine the time t at which these meet.

Vf = V0 + at

In addition, I chose to set up the problem such that negative acceleration was down (ie. g = -10 m/s^2)

For rock A (the one being dropped):
Vf = V0 + at becomes Vf = 0 - at which becomes Vf = -10t(notice Vf is negative)

For Rock B (the being thrown up)
Vf = V0 + at becomes Vf = V0 - at, but now we need V0.

well, in a projectile problem, the V0 of the upward trajectory is equal to the Vf of the downard portion of the problem, so V0 (of B) is equal to Vf (of A) which is nothing but sqrt (2gh) (from vf^2 = v0^2 + 2ax for free fall). I arrived at V0 = 20*sqrt(5) = approx. 44.7

so, for Rock B:
Vf = V0 - at becomes Vf = 44.7 - 10t
EQUATING THE TWO Vf's gives:

-10t = 44.7 - 10t

well, the -10t's seem to cancel, BUT, IN FACT THEY ARE POINTED THE OPPOSITE WAYS.....so to make them truly equal, you have to add a negative....hence,

10t = 44.7 - 10t which gives t = 44.7 / 20 = 2.235 seconds.

with t = 2.235 just plug and chug in to rock a or b ( I chose a):
x = v0t - 1/2 at^2 which gives -25 meters (with v0 = 0).

or just 25 meters from the top of the building down....or 75 meters.

Hope that made sense, carolinamcat.

Excellent explanation Bigserve. Might I add a little bit that may make the question go a bit faster.

The key here, as Bigserve pointed out, is that they each travel the same distance, but in opposite directions. Assuming the each travel for the same time (which they must to cover the same distance), we can use 1/2(10)texp2 = 100 to solve for the total travel time (about 4.5 seconds). This means that the half time is about 2.25 seconds. At 2.25 seconds after the start of the experiement, the two balls pass one another. From there, it's a matter of plugging into the same equation again to determine how far the dropped object has fallen in 2.25 seconds.

As a point of interest, an object in free fall travels three times the distance in its second period of time than its first half of time. For instance, had the building been 200m, then the meeting point would have been 150m and so on.

You could also have solved this intuitively (through visualization). In order for the ball to travel up to the top of the roof, it must be launched with a significant vertical velocity. Over the duration of its flight, it will slow down. For the dropped ball, it will start slowly and pick up speed as it falls. Assuming they each travel the same distance in the same time (if they didn't, then the launched ball would not travel 100 m), then we can compare their paths directly. The dropped ball covers less distance in the first half (timewise) of its jouney while the launched ball covers more distance in the first half (timewise). This means that they must intersect at a point above halfway distance wise. The answer must be above 50m and below 100m. I have no idea what the answer choices are, but you should eliminate any answers outside that range and be pulled to a value close to the middle of the range (about 75m). This intuitive way is fast, but will likely leaving you feeling uncertain. It's a personal choice you need to make whether speed or certainty is more important to you.

 

[carolianamcat]

hi bigserve99 and berkreviewteach,
Thank you so very much for taking the time to explain this to me. It all makes sense now. You just have to look at what is common in between both the cases. The only question I have is - why is vf negative?
For rock A (the one being dropped):
Vf = V0 + at becomes Vf = 0 - at which becomes Vf = -10t(notice Vf is negative)
Isn't acceleration in the same direction as the velocity in this case?
Thanks.

 

[bigserve99]

hi bigserve99 and berkreviewteach,
Thank you so very much for taking the time to explain this to me. It all makes sense now. You just have to look at what is common in between both the cases. The only question I have is - why is vf negative?
For rock A (the one being dropped):
Vf = V0 + at becomes Vf = 0 - at which becomes Vf = -10t(notice Vf is negative)
Isn't acceleration in the same direction as the velocity in this case?
Thanks.

Hi Carolina,

Yes, acceleration is in the same direction as Vf for the rock A. However, the system that one chooses to set up, namely which direction is positive and negative, is universal to both acceleration and velocity. In other words, all velocities and acceleration in the downward direction is negative. They are not relative to one another. When rock B is thrown up.....the acceleration is still negative, because it points down, but now the velocity is positive because it points up. With rock A, velocity is downard along with accleration, which is making it more negative, hence Vf for Rock A is negative.

EDIT: You will also notice my final vertical displacement is -25 m (negative). It's not just acceleration and velocity that is universally negative or positive, so is displacement.

Hope that helps.

Thanks BerkReviewTeach for the extra insight, much appreciated.

 

[Pietrantonio]

Just wondering if anyone could help me solve this problem. I have a general idea, however, my answer doesn't match with the correct answer.

Specifically deals with Newtons Second Law

Question:
A jet catapult on an aircraft carrier accelerates a 2000-kg plane uniformly from rest to a launch speed of 320 km/h in 2.0s. What is the magnitude of the net force on the plane?

Answer Provided:
8.9x10^4 N

I appreciate your help very much.

 

[bigserve99]

Just wondering if anyone could help me solve this problem. I have a general idea, however, my answer doesn't match with the correct answer.

Specifically deals with Newtons Second Law

Question:
A jet catapult on an aircraft carrier accelerates a 2000-kg plane uniformly from rest to a launch speed of 320 km/h in 2.0s. What is the magnitude of the net force on the plane?

Answer Provided:
8.9x10^4 N

I appreciate your help very much.

Sure,

So, it's a kinematic problem combined with F=ma:

First, calculate the acceleration required to achieve the Vf, 320 km/h which is approx. equal to 1 x 10 ^ 2 m/s (if you need help with this conversion, let me know).

Vf = V0 + at

1x10^2 (m/s) = 0 + a * (2 seconds)

a = 0.5 x 10^2 (m/s^2)

plug in a in F=ma:

F = 2x10^3 (kg) * 0.5 x 10^2 (m/s^2) = 1 x 10^5 N

1 x 10^5 N is approx. equal to your answer 8.9x10^4. I just used approx. throughout the problem. If you do the calculations, espeically using the exact ratio of 3.2/3.6 for finding Vf, you'll find the more exact answer you have given; however, I doubt you would need it that accurate for the MCAT.

BTW, I'm not even sure I'm allowed to post here as far as helping people with answers. Is this ok? In any case, if anything is wrong with my work, please let me know anyone.

Hope that helped.

~ BigServe99

 

[gridiron]

Sure,



BTW, I'm not even sure I'm allowed to post here as far as helping people with answers. Is this ok? In any case, if anything is wrong with my work, please let me know anyone.

Hope that helped.

~ BigServe99

You are allowed to post here to help people. There is nothing wrong with your work and your input is greatly appreciated.

BTW, I hope the Indians go all the way----this coming from a white sox fan

 

[Pietrantonio]

BigServe99,

Thanks for helping me with the problem. I fully understand the method to solve this problem now, however, I am going to have to take you up on the offer of you explaining the conversion of 320km/h to 1x10^2 m/s. What I had tried before posting this problem, my conversion factor looked like this:

320 km / h (1 h/ 60 min) (1 min / 60 sec) (1 m / 10^3 km) = ~ 8.9x10^-5

I feel that something has gone wrong with the conversion factor. If you could explain, I'd greatly appreciate it. Thanks !

 

[Pietrantonio]

Actually I think I figured it out. I'm assuming I went wrong with the (1 m/ 10^3 km) I think the conversion should be flipped to (10^3 m / 1 km). So, this should give me a final result of ~8.9x10^1

Then, I used the formula Vf = V0+at

8.9x10^1 = 0 + a *(2s)
a= 4.45x10^1

Pluging a into F=MA like you said,

F= (2x10^3 kg) * (4.45x10^1 m/s^2) = 89,000 or 8.9x10^4N

If this happens to be wrong, please let me know, but I think I have figured this out with your help and advice.

 

[Pietrantonio]

As if I have already given enough to do, I happened to come across another problem that I don't understand. It involves projectile motion.

Question:
A ball rolls horizontally with a speed of 7.6 m/s off the edge of a tall platform. If the ball lands 8.7 m from the point on the ground directly below the edge of the platform, what is the height of the platform?

Answer:
6.4 m

My synopsis:
The information seems fairly limted, but I am assuming that we need to find t (time in seconds), and then put all of the infomation into a kinematic equation to solve for the height of the platform. I appreciate all of your help!

 

[bigserve99]

As if I have already given enough to do, I happened to come across another problem that I don't understand. It involves projectile motion.

Question:
A ball rolls horizontally with a speed of 7.6 m/s off the edge of a tall platform. If the ball lands 8.7 m from the point on the ground directly below the edge of the platform, what is the height of the platform?

Answer:
6.4 m

My synopsis:
The information seems fairly limted, but I am assuming that we need to find t (time in seconds), and then put all of the infomation into a kinematic equation to solve for the height of the platform. I appreciate all of your help!

Yes, that is correct. We need to find the time of flight. Based on this, it is a kinematic problem utilizing x = v0*t + (1/2)*a*t^2.

Basically the problem is split into the horizontal components and vertical components. The horizontal velocity is constant throughout the problem and is so during any uniform projectile problem in which the net external forces are balanced in the horizontal direction. So, V0x * time of flight (t) = x (displacement in the horizontal direction). This displacement is also called the "range" in a lot of problems. In any case, in this problem, V0x and x is given. So,

7.6 (m/s) * t (s) = 8.7 (m) or t = 8.7/7.6

(NOTE: I attempt all these problems without a calculator for the MCAT's sake)

I approximate t as 1.13 - I do this by subtracting 7.6 from 8.7 to give 1.1 and divide the 1.1 over 7.6 for an approximate decimal value. I rounded 1.1 to 1 and 7.6 to 8. so 1/8 is .125 or just a quarter of a quarter.

using 1.13 for my time, just plug and chug in y = (1/2) * a * t^2. using -10 (m/s^2) for a.

So, y (the displacement from the top of the building) = 0.5 *(10) * 1.13^2

To approx. 1.13^2, I break up in my head (1.13 * 1.13) into 1.13 * (1 + 0.13) which gives 1.13 + 1.13*(0.13) which is just 1.13 + .13 + (0.13*0.13). At this point I just ignore the 0.13*0.13 (it's a small number). Plus by now I'm sure we could find the answer out of the choices give. I mean, the t^2 I approx. to about 1.26 or almsot 1.3 based on what I jsut wrote: 1.13 + 0.13 (leaving off 0.13 *0.13). 1.3 * 5 is easy enough...to give approx. 6.5

Realizing I've been rounding up in most situations, tells me my answer is a little bit below that.

BTW, I know that it sounds insane the way I break up the approximations, as far as how long it is to write out, but honestly, I do that with every problem in my head and now those approx. take but a few seconds. I mean I even do it in lab, anywhere I need to calculate stuff that doesn't need to be exact.

Hope that helped,
~ Bigserve99




Oh and gridiron, thanks for the vote of confidence in the Indians.....people always overlook the AL Central, and look what's happened the past few years.

 

[Pietrantonio]

Awesome! Thank-you BigServe99. Your help is appreciated very much. Just a question about physics. Is it just repetition that allows you to know eactly what you need to find and what equations to use, or is there something in every problem that sort of hints at what needs to be done? I ask, because I have a hard time understanding what equation to use, and what I need to find first. I'm sure it just takes lots of practice, but Physics seems to be my weakest area. Thanks for all of your help!

 

[Pietrantonio]

The only question I have at the moment for the last question I asked, is the formulas you used to solve it.

Solving for t
Is this the standard equation used when you always need to find t?
x=V0x * t

Solving for the height of the building y
Is this the standard equation used to always find y?
y=0.5at^2

For some reason, I'm having a hard time with kinematic equations.