In the figure, a small block of mass m 0.037 kg can slide along the frictionless

Question

In the figure, a small block of mass m 0.037 kg can slide along the frictionless loop-the-loop, with loop radius R = 19 cm. The block is released from rest at point P, at height h = 9R 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?

Explanation / Answer

Mass of the block is, m = 0.037 kg

Radius of the loop, R = 19 cm = 0.19 m

h = 9R

g = 9.8 m/s2 is the acceleration due to gravity.

(a)

Work done by the gravitational force is in moving the block from point P to point Q is,

W = - (mgR - mgh)

W = - 0.037 * 9.8 * (0.19 - 9* 0.19)

W = 0.037 * 9.8 * 8 * 0.19

W = 0.551152 J = 0.55 J

(b)

Work done by the gravitational force is in moving the block from point P to top of the loop is,

W = - (mg2R - mgh)

W = - 0.037 * 9.8 * (2*0.19 - 9*0.19)

W = 0.037 * 9.8 * 7 * 0.19

W = 0.482258 J = 0.48 J

Gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop.

(c) Potential energy of the block at point P:

U = mgh

U = 0.037 * 9.8 * 9 * 0.19

U = 0.620046 J = 0.62 J

(d) Potential energy of the block at point Q:

U = mgR

U = 0.037 * 9.8 * 0.19

U = 0.068894 J = 0.07 J

(e) Potential energy of the block at the top of the loop:

U = mg(2R)

U = 0.037 * 9.8 * 2 * 0.19

U = 0.137788 J = 0.14 J

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