AAMC Self-Assessment Physics #96

[DSimone]

Q: How will W [work done by friction] change if the initial speed of the box at Point A is increased by a factor of 2?

A-W will decrease by 50%
B-W will not change
C-W will increase by 50%
D-W will increase by 100%

I feel like this should be very simple, but I am hung up on this problem for some reason . I understand that given the fact that W=Fd cos the angle of incline and that Ff is independent of velocity (because it is determined by normal force and the coefficient of friction), the work done by friction will not change.

But, I also thought that work done by friction was related to the change in KE, which IS dependent on velocity, no? I am not clear why additional work is not done by the force of friction if the initial speed is increased. Won't the overall change in KE be greater with a greater initial velocity?

Essentially, my question is - how are the equations W=Fd and W=(delta KE + delta PE) related if one is independent of velocity and the other is not? Could someone kindly shed some light on this for me? Thanks!

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[polyatomic]

It's a little difficult to approach this problem from the standpoint of kinetic energy and velocity. It's easier if you look at it from the potential energy.

So Work = ΔKE = ΔPE

Moving down the slope, no matter what speed, we'll always have the same change in height, thus the same ΔPE, thus the same ΔKE, and finally the same work done by gravity. So it makes sense that friction would also perform the same amount of work, irrespective of velocity.


This is different when we look at problems where we travel linearly. For example, if a car is traveling 10 m/s and stops in 10 m, we find work by doing: W = ΔKE = (10^2)(10) = 1,000 J.

When we double the speed here, we're quadrupling the W = ΔKE = (20^2)(10) = 4,000 J.

In this case, speed DOES change the work done. But remember that in the linear case, we're changing the difference in velocity for both cases from start to finish.

So I suppose the reason why increased speed is not changing the work done in your inclined plane problem is because the velocity is changing proportionally so the ΔKE always remains the same, matching the ΔPE.

 

[DSimone]

Ahhh-ha!

Thanks, polyatomic, for your insight! I was not accounting for the fact that there is an initial KE (since we are not descending the slope from rest). As you stated in your example, in many linear cases we are looking at a situation in which work = ΔKE from start to finish (either beginning from or ending at rest) and that is not the case here.

Does this sound right?...

In this problem, Work=KE1+PE1=KE2+PE2 --> Work=PE1=KE2-KE1. So, regardless of the initial and final velocities work will always be equal to PE1.

Initially, I was assuming the final velocity at the bottom of the slope would be the same both before and after doubling the initial velocity because the problem states that final KE=mgd. But, it appears that while initial velocity will change, final velocity will also change - proportionally - so that ΔKE is the same in both cases. I did not account for this in my original approach to the problem.

I understand that if ΔKE does not change, work done by friction will not change. But I am still a bit confused about the statement in the problem that the box "reaches point B with a speed of vB and kinetic energy equal to Mgd." If vA is doubled, won't vB also be greater? If so, how can the final KE both before and after doubling the initial velocity still equal Mgd? Wouldn't final KE with the doubled initial velocity also be greater? Or, am I totally off here???

 

[DSimone]

Anyone? [bump]

 

[sdm33]

I hate this topic , can someone break it down for me .Thanks

If we say that W = F.d = m.a .d = (m.V.d)/t
if V double then W double. Can we say that or no????

Two identical jobs or tasks can be done at different rates - one slowly or and one rapidly. The work is the same , does that mean work is not depended on velocity ?


If KEi increases due to velocity increase , but the work is due to friction is decreasing the KEi so do they balance out???

 

[JLeBling]

I'm having the same issue. This was my (INCORRECT) line of reasoning:

w= work
W=work by friction
w = Fd
v = d/t

By increasing the (v) by a factor of 2, (d) increases by a factor of 2.
Which would give us a value of 2w.
And since W = -w, it follows that W would increase by 100%.

BUT the question is limiting the frame of reference to the "work done by friction on the box as it travels down the ramp which is given by W". So I am wrong in assuming that we are considering work done after the box has slid off the ramp past Point B and E and C.

So I guess the appropriate way to look at it is W = -Fd.
The since the length of the ramp does not change, (d) does not change.
(F) does not change either since F = ma = mg
The mass of the box has not changed. The acceleration due to gravity hasn't changed either.
SO HOW THE HECK WOULD THE WORK DUE TO FRICTION CHANGE?

You get greased if you miss the definition of W! Don't get greased!

 

[Postictal Raiden]

Here's what I think:

W(friction) = d x F(friction) = d x N x friction coefficient = d x mgcos x u

d and m don't change. the acceleration also remain the same regardless of the initial velocity. It's like comparing the acceleration between an object that is dropped form a given height to an object that was thrown down from the same height. Both will have the same acceleration regardless. All the variables in the equation are fixed, so work done by friction remains constant.

 

[Meredith92]

It's a little difficult to approach this problem from the standpoint of kinetic energy and velocity. It's easier if you look at it from the potential energy.

So Work = ΔKE = ΔPE

Moving down the slope, no matter what speed, we'll always have the same change in height, thus the same ΔPE, thus the same ΔKE, and finally the same work done by gravity. So it makes sense that friction would also perform the same amount of work, irrespective of velocity.


This is different when we look at problems where we travel linearly. For example, if a car is traveling 10 m/s and stops in 10 m, we find work by doing: W = ΔKE = (10^2)(10) = 1,000 J.

When we double the speed here, we're quadrupling the W = ΔKE = (20^2)(10) = 4,000 J.

In this case, speed DOES change the work done. But remember that in the linear case, we're changing the difference in velocity for both cases from start to finish.

So I suppose the reason why increased speed is not changing the work done in your inclined plane problem is because the velocity is changing proportionally so the ΔKE always remains the same, matching the ΔPE.

I'm still pretty confused by this problem.
Polyatomic, you said: So Work = ΔKE = ΔPE but I thought Work = change in KE + Changing PE. Am I wrong on this?

DSimone, this is starting to make a little more sense when you wrote:

In this problem, Work=KE1+PE1=KE2+PE2 --> Work=PE1=KE2-KE1. So, regardless of the initial and final velocities work will always be equal to PE1.

But I again get confused after reading this:

I understand that if ΔKE does not change, work done by friction will not change. But I am still a bit confused about the statement in the problem that the box "reaches point B with a speed of vB and kinetic energy equal to Mgd." If vA is doubled, won't vB also be greater? If so, how can the final KE both before and after doubling the initial velocity still equal Mgd? Wouldn't final KE with the doubled initial velocity also be greater? Or, am I totally off here???


can someone clarify this??

Do you think they meant to say: "Reaches point B with a speed of Vb and kinetic energy CHANGE equal to mgd?" I think this makes more sense. Kinetic energy would only equal mgd if kinetic energy was initially zero and then it gains what was lost from potential E

 

[Meredith92]

I also got question 99 wrong, which uses the same concept. "How will W change if the angle of the ramp to the horizontal is increased?"

I thought that if the angle was increased, the change in potential E would be greater (since youre starting it off at a higher height) and since W of a nonconservative force =change in PE + change in KE then work would increase.

however, they said that "W will decrease because the normal force to the surface of the ramp will decrease"

I always understand the answer when they put it in terms of the W equation (which doesnt include velocity/PE/KE) but when I try to reason out these types of problems in terms of KE and PE i always get them wrong!! Can someone help me out to show me what I am misunderstanding??

Thanks so much for your help!

 

[Postictal Raiden]

I also got question 99 wrong, which uses the same concept. "How will W change if the angle of the ramp to the horizontal is increased?"

I thought that if the angle was increased, the change in potential E would be greater (since youre starting it off at a higher height) and since W of a nonconservative force =change in PE + change in KE then work would increase.

however, they said that "W will decrease because the normal force to the surface of the ramp will decrease"

I always understand the answer when they put it in terms of the W equation (which doesnt include velocity/PE/KE) but when I try to reason out these types of problems in terms of KE and PE i always get them wrong!! Can someone help me out to show me what I am misunderstanding??

Thanks so much for your help!

I'm going to go on a limb her and say, I think this is the downfall of exclusively using TBR for your prep. TBR goes way beyond what you need to know on the mcat. Therefore, when confronted with an easy straightforward question, your this-is-a-trick-question alarm goes off.

Just look at the equation of work done by friction, W = F(friction) x d, then ask yourself what happens when to work when the angle increases. We all know that as the angle increases, the value of costheta decreases, which in turn decreases the normal force and thus the frictional force. Therefore, W decreases.

 

[Meredith92]

You're definitely right. I see it now as a very easy question. But I guess I still want to understand it in terms of energy because that always made sense to me in physics class and it's frustrating that I can't reason it with KE and PE anymore:/ if anyone has a way to think of it in terms of energy tht works I'd love to hear it.

But thanks Ibn for your honesty, I def need to be careful to not over think things

 

[Lolaaa]

maaaaan this is so confusing. W= delta KE is throwing me off. if we double the initial velocity doesn't that increase the mag of delta KE thus increasing the mag of W? Somebody here mentioned that when the initial velocity changes, the final velocity also changes and I don't get that. I know this is an old thread but please somebody explains to me what I'm missing

 

[Cornfed101]

maaaaan this is so confusing. W= delta KE is throwing me off. if we double the initial velocity doesn't that increase the mag of delta KE thus increasing the mag of W? Somebody here mentioned that when the initial velocity changes, the final velocity also changes and I don't get that. I know this is an old thread but please somebody explains to me what I'm missing

A lot of energy is lost to heat with friction. In other words PE isn’t all converted to KE. The relationship falls apart here.

 

[Lolaaa]

A lot of energy is lost to heat with friction. In other words PE isn’t all converted to KE. The relationship falls apart here.

ohhh that makes sense now. Thank you!

 

[PlsLetMeIn21]

This question is really typical of what they like to do. They give you an extraneous factor to draw your focus.

As long as the object makes it to the same point in each trial, then the work done by friction will be mukmgcos(theta) no matter how fast it goes or how long it takes. The deltaPE and delta KE will be the same in each case (fast or slow), so the work done by friction is the same.

This question is egging you on to overthink it. Look for that on the real deal too.