PE & KE Problem: Work, PE & KE Calcs

[vysero]

In Fig. 8-31, a small block of mass m = 0.032 kg can slide along the frictionless loop-the-loop, with loop radius R = 12cm. The block is released from rest at point P, at height h = 5.0R above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point P to (a) point Q and (b) the top of the loop? If the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point P, (d) at point Q, and (e) at the top of the loop? (f) If, instead of merely being released, the block is given some initial speed downward along the track, do the answers to (a) through (e) increase, decrease, or remain the same?
I believe that the relevant equations are:
w=fd and w=mgh and maybe ke=1/2mv^2
Well I believe if I could figure out what the the specific heights at the given points are then I could make some calculations for instance: (a) If I new what the height at point Q was I would say the work would be mgh but I am having trouble understanding what they want me to do with the given height being 5.0R! Also, I believe the answer to (f) is remain the same right because work and PE don't have anything to do with velocity.

 

[haruspex]

The heights of points P and Q are clearly given in the diagram: h and R respectively, and you know h = 5R.
Q appears to be intended as the height of the centre of curvature of the final loop. The radius is given as R, so what would the height of the top of the loop be?

 

[vysero]

Well let's start with the height at Q. The height at is 5R or 5*12cm which is 60cm. So then point Q is 60cm - 12cm which is 48cm right? That's .48m and if w = mgh then it should be (.032)(9.8)(.48) = .150528 J which doesn't seem right to me.

 

[haruspex]

Well let's start with the height at Q. The height at is 5R or 5*12cm which is 60cm. So then point Q is 60cm - 12cm which is 48cm right? That's .48m and if w = mgh then it should be (.032)(9.8)(.48) = .150528 J which doesn't seem right to me.

No, P is 5R above the bottom of the loop, and Q is R above the bottom of the loop. Q is 48cm below P, but I think they want the PE relative to 0 at the bottom of the loop.

 

[Nickie]

I would first clarify the problem by asking for the specific values for the heights at points Q and the top of the loop. Without this information, it is difficult to accurately calculate the work done and potential energy at these points.

Assuming that the height at point Q is 4.0R (based on the given height of 5.0R at point P), the work done by the gravitational force on the block as it travels from point P to point Q would be w = mgh = (0.032 kg)(9.8 m/s^2)(4.0R) = 1.26 J.

The work done as the block reaches the top of the loop would depend on the final height, which is not specified in the problem. However, it can be assumed that the work done would be the same as the work done at point P, since the block is released from rest at both points and the only force acting on it is gravity.

If the gravitational potential energy is taken to be zero at the bottom of the loop, then the potential energy at point P would be mgh = (0.032 kg)(9.8 m/s^2)(5.0R) = 1.56 J. At point Q, the potential energy would be mgh = (0.032 kg)(9.8 m/s^2)(4.0R) = 1.26 J. And at the top of the loop, the potential energy would depend on the final height, which is not specified.

If the block is given some initial speed downward along the track, the answers to (a) through (e) would remain the same. As you mentioned, work and potential energy do not depend on the velocity of the object, only on the height. However, kinetic energy would increase as the block gains speed, and this would need to be taken into account in future calculations.