TBR Physics I, Work/Energy Passage Question

[arc5005]

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In order to move boxes with a mass between 80-100 kg, a storage facility set up a lift, frictionless ramp, and roller. The lift contained a flat metal lift plate onto which a box could be placed. It is connected to a counterweight by vertical cables around a pulley wheel hanging from the ceiling. A mass of 50 kg was chosen for the counterweight.

***There's only this 1 image, I dunno why the x is showing up below it
IMG 2826

Increasing theta while keeping point a at the same spot would have what effect on the apparatus?
I. The speed of the box at point b would increase.
II. The work done by friction in the region from point b to point c would increase.
III. The number of rollers needed between point b and point c would remain the same.


A. III only
B. I and III only
C. II and III only
D. I, II, and III

A) III only
If point a remains at the same spot, then the height at the top of the ramp remains the same. This means the box starts its descent with the same initial PE. As a result, it will have the same KE at the bottom of the ramp. The mass of the box remains the same, so the maximum speed of the box must also remain the same. By increasing the angle, the slope of the ramp is steeper, which means that the acceleration is greater (a = g sinθ). But, it travels a shorter distance, so it is accelerated for a shorter amount of time, so it reaches the same speed at the bottom of the ramp (point b). Statement I is invalid, which eliminates choice B and D, so now Statement III must be true, because it's in both choices A and C. The work done by friction is equal to the change in KE between points b and c, which happens to equal the change in PE from point a to point b. This means that Statement II is invalid, and choice A is the best answer.

Question:
Having a hard time grasping and understanding this question, and why each statement is either true or false. Can anyone help clarify?

Okay, so they are saying by increasing theta, we are making the angle in reference to the horizontal larger (the ramp steeper), that means that when theta = 90 degrees, then the object would technically be in free fall?

So because the ramp is frictionless no energy is being lost to friction, so PEinitial = KEfinal? They're saying because the mass of the box stays the same, then the speed has to stay the same, because KE has to = PE, and since PE is not increasing, then KE is not increasing?

PE = mgh or PE = mgxsinθ

PE = (50)(10)(h)

KE = 1/2mvf^2 - 1/2mvi^2

KE = 1/2(50)vf2 - 0

PE = KE

(50)(10)(constant) = 1/2(50)vf2 ???? would it be similar to this? h doesn't change, but xsintheta would change, so I guess i'm a bit confused.

 

[AdaptPrep]

Greetings, arc5005.

You are understanding the problem correctly in that theta increases (the base of the triangle gets shorter).

This problem can be thought of conservation of energy. You are right, all the potential energy is being changed to kinetic energy since there is no friction. Since the height (point A) does not change, the PE does not change. Hence, KE does not change either...the velocity is always the same. Hence, you can eliminate option I. And, because you have the same velocity, no more work needs to be done to stop it (energy change is the same in all circumstances), so option II can be eliminated. Option III implies that the same amount of energy is involved in stopping the box (remember, delta KE is the same in all situations where height is constant), and is therefore, the correct answer.

If you are wondering what changes with changing theta (angle), two things change. The rate of acceleration that the box has going down the slope will change. However, the amount of time also changes. For example, a smaller theta (less incline) will mean slower rate of acceleration but a longer time that it will accelerate. A larger theta (more incline) will mean greater rate of acceleration but a shorter time that it will accelerate. They actually will always give the same final velocity.

I hope that helps!

Good luck studying! Physics can be tricky at times.

 

[DPTinthemaking15]

Hey @AdaptPrep I have a question about this problem that you may be able to answer. The answer explanation states "But, it travels a shorter distance, so it is accelerated for a shorter amount of time, so it reaches the same speed at the bottom of the ramp (point b)." How does it travel a shorter distance if everything stays the same, except for increasing theta?

 

[AdaptPrep]

Hi, DPTinthemaking15--

If you increase theta, the hypotenuse has to get shorter. For example:

The height was kept constant (4 m), but increasing theta will shorten the hypotenuse. The box slides down the hypothenuse, so the distance traveled by the crate is shorter. Hence, it has less time to accelerate. It will accelerate quicker but for a shorter period of time resulting in the final velocity at point b being the same.

I hope that answers your question. If not, let me know.

 

[DPTinthemaking15]

Hi, DPTinthemaking15--

If you increase theta, the hypotenuse has to get shorter. For example:
View attachment 234787

The height was kept constant (4 m), but increasing theta will shorten the hypotenuse. The box slides down the hypothenuse, so the distance traveled by the crate is shorter. Hence, it has less time to accelerate. It will accelerate quicker but for a shorter period of time resulting in the final velocity at point b being the same.

I hope that answers your question. If not, let me know.

THANK YOU!!! As usual, you are a life saver on these forums!

 

[PlsLetMeIn21]

That is a super explanation Adapt, thank you.