Little gravity; stopping distance (friction)

[Bond03]

---

Members do not see this ad.

The question is as follows (from Berkley Review Physics)

After sliding down the slide, a child then travels across the rough surface until coming to rest. Which of the following equations best represents how far the child will travel before stopping?

Correct answer: s=h/muK
Other answer: s=mgh/muK

I understand why the correct answer is the correct answer but I dont get why the other answer is incorrect. The explanantions goes something along the lines that if gravity decrease, stopping distance should increase. WHY??? Maybe I dont fully understand gravity 🙁 😕

Thanks in advance

 

[Dr Gerrard]

I am not sure what that explanation is all about. It seems as though, according to the explanation, the stopping distance does depend upon g, but according to the solution they chose, the stopping distance is independent of g.


I tried doing this problem out to try and help, but I can't even get the correct answer.

I tried using the work energy theorem.

deltaK = net Work = Wg + Wf

Where Wg is work done by gravity and Wf is work done by friction

deltaK is 0 here, since that is the final and initial velocity.

Thus, Wg = Wf (since the work done by friction is negative compared to the work done by gravity).

mgh = N * muk * s where N = normal force = mgcos(theta), where theta is the angle between the slide and the ground, and where s = distance the child slides.

SO --- mgh = mgcos(theta) * muk * s

h = cos(theta) * muk * s

S = h/cos(theta)*muk*s

See, I get the same answer with the cos(theta) on the bottom that comes from the normal force. Where did I go wrong?

 

[johndoe3344]

I didn't get a chance to read your work, Dr Gerrard, but here is the solution:

Use the kinematics formula: Vf^2 = Vo^2 + 2a*x.

Solve for x, with Vf=0 ==> x = -Vo^2/(2a)

Realize that at the top of the slide, the PE is max, and at the bottom of the slide all of that PE is converted to KE. We have mgh = 1/2*m*v^2. This v is the Vo in the above formula.

Realize that the "acceleration" is the stopping force, which in this case, is the force due to kinetic friction, with a = -Fk = -uK*m*g

Plug everything in and we get x = -2*g*h / -2*uK*m*g which simplifies to h/m*uK as claimed.

 

[Dr Gerrard]

John Doe,

You come up with the right answer, but I am not seeing it.

You say that at the top, PE is max and at the bottom all the PE is converted to KE, making KE max at the bottom.

However, you say that Vf at bottom = 0 which would make kinetic energy at the bottom equal to 0.

Additionally, you say that Vi at the top is equal to the same V in your final kinetic energy equation, even though Vi is also equal to 0, since the child starts from rest.

So I basically did teh same thing as you, but rather than saying mgh = KE, I set mgh = work done by friction, since eventually I said that all of the potential energy goes into friction, since the initial and final velocities are equal to 0.

 

[johndoe3344]

I think you're misunderstanding a few things.

It is correct that PE is max at top, and all the PE is converted to KE.

But I used Vi and Vf to denote the initial and final velocities of the child on the rough surface. Therefore, Vi would be the v in the KE formula, while Vf would be zero because the child eventually stops on the rough surface.

Does this make it clearer?

 

[Dr Gerrard]

I get the answer now, thanks John Doe.

Anyways OP, gravity does not affect the stopping distance. I am not sure why they wrote that as a solution. But if gravity were to decrease, the child would have less potential energy at the top, and the child would have less kinetic energy at the bottom. However, the child would also feel a smaller force due to friction, since the normal force at the bottom would also be smaller. Thus, the effect of gravity would cancel out, as made evident by the correct answer choice.

Sorry for hijacking your thread as well. Hope my above explanation helps.