Question

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* = 5*R* 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?

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

Step 3

Part B:

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

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