In the figure below, a small block of mass m = 0.013 kg can slide along the fric
ID: 1541812 • Letter: I
Question
In the figure below, a small block of mass m = 0.013 kg can slide along the frictionless loop-the-loop, with loop radius R = 13 cm. The block is released from rest at point P, at height h = 5.1R above the bottom of the loop.
(a) How much work does the gravitational force do on the block as the block travels from point P to point Q?
(b) How much work does the gravitational force do on the block as the block travels from point P to 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, find the following.
(c) the potential energy when the block is at point P
(d) the potential energy when the block is at point Q
(e) the potential energy when the block is at the top of the loop
(f) If, instead of 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?
In the figure below, a small block of mass m 0.013 kg can slide along the frictionless loop-the-loop, with loop radius R 13 cm. The block is released from rest at point P, at height h 5.1 m above the bottom of the loop. (a) How much work does the gravitational force do on the block as the block travels from point Pto point Q? (b) How much work does the gravitational force do on the block as the block travels from point P to 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, find the following. (c) the potential energy when the block is at point P (d) the potential energy when the block is at point Q (e) the potential energy when the block is at the top of the loop f) If, in of 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? stead increase O decrease remain the sameExplanation / Answer
Height of point P in meters.
h = 5.1 * 0.13 = 0.663 m
block’s potential energy at point P.
PE = 0.013 * 9.8 * 0.663= 0.0844662 J
block’s potential energy at point Q. At point Q, its height is 0.13 m.
PE = 0.013 * 9.8 * 0.13 = 0.016562 J
(a)
The work is equal to the difference of the two potential energies.
Work = 0.0844662 – 0.016562 = 0.0679042 N * m
(b)
At the top of the loop, the block’s height is 0.26 m.
PE = 0.013 * 9.8 * 0.26 = 0.033124 J
The work is equal to the difference of the block’s initial potential energy and this number.
Work = 0.0844662 – 0. 033124 = 0.0513422 N * m
(c)
PE = 0.0844662 J
(d)
PE = 0.016562 J
(e)
PE = 0.033124 J
(f) Remain the Same
Potential energy and work doesn’t depend on speed