Question

A small block of mass m = 0.032 kg can slide along the frictionless loop-the-loop, with loop radius R = 12 cm. 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? (0.15 J)
b) the top of the loop? (0.11 J) 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? (0.19 J) d) at point Q? (0.038 J) e) at the top of the loop? (0.075 J) f) If, instead of merely being released, the block is given some initial speed downward along the track, do the answers to a) through c) increase, decrease or remain the same? (same)

Answer 1

Answer:

Part a)

$W = 0.15 J$

Part b)

$W = 0.11 J$

Part c)

$U = 0.19 J$

Part d)

$U = 0.038 J$

Part e)

$U = 0.075 J$

Part f)

It is independent of the speed of the object so all part answers will remain the same

Explanation:

Part a)

As we know that Point P is at height 5R while point Q is at height R

so the work done by gravity from P to Q is given as

$W = mg(5R - R)$

$W = 0.032(9.8)(4)(0.12)$

$W = 0.15 J$

Part b)

When it reaches to the top of the loop then its final height from ground is

h = 2R

so work done from P to Q is given as

$W = mg(5R - 2R)$

$W = 3mgR$

$W = 0.11 J$

Part c)

Potential energy at P point is given as

$U = mgH$

$U = 0.032(9.8)(5)(0.12)$

$U = 0.19 J$

Part d)

Potential energy at Q point is given as

$U = mgH$

$U = 0.032(9.8)(0.12)$

$U = 0.038 J$

Part e)

Potential energy at top point is given as

$U = mgH$

$U = 0.032(9.8)(2)(0.12)$

$U = 0.075 J$

Part f)

Since all the answer from part a) to part e) depends only upon the position of the object.

So here we can say that it is independent of the speed of the object so all part answers will remain the same