In the figure, a small block of mass m = 0.047 kg can slide along the frictionless loop-the-loop, with loop radius R = 20 cm. The block is released from rest at point P, at height h = 5R 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?

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Asked Mar 25, 2019
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In the figure, a small block of mass m = 0.047 kg can slide along the frictionless loop-the-loop, with loop radius R = 20 cm. The block is released from rest at point P, at height h = 5R 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?

 

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Expert Answer

Step 1

Given mass of the block m = 0.047 kg

consider acceleration due to gravity g as 10 m/s2.

R = 20 cm = 0.2 m

h = 5R = 100 cm = 1m

work done by the gravitational force  = ( Gravitational Force )* (displacement in the direction of gravitational force )

work done by the gravitational force  = (mg)*(displacement)

Step 2

PART A:

Work done(WD)  by the gravitational force on the block as the block moves from point P to point Q

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Step 3

Part B:

Work done(WD)  by the gravitational force on the blo...

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