In the figure, a small block of mass m = 0.012 kg can slide along the frictionle

Question

In the figure, a small block of mass m = 0.012 kg can slide along the frictionless loop-the-loop, with loop radius R = 17 cm. The block is released from rest at point P, at height h = 4R 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?

In the figure, a small block of mass m = 0.012 kg can slide along the frictionless loop-the-loop, with loop radius R = 17 cm. The block is released from rest at point P, at height h = 4R 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

Given

block of mass m = 0.012 kg

loop radius R = 0.17 m

height h = 4R

Acceleration due to gravity g = 9.8 m/s2

When the block reaches from P to Q,

     It covers heighth ' =h -R

                                  = 4R - R

                                 = 3R

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a) work does the gravitational force do on the block asthe block travels from point P to point Qis

             W = mgh'

                   =mg ( 3R)

                 = 3 (0.012 kg )( 9.8m/s2)( 0.17 m) = 0.06 J

b) work does the gravitational force do on the block as theblock travels from point P to the top of the loop is

       W = mgh''

Here h'' = 4R - 2R

              = 2R

      W = 2 mgR

          =2 (0.012 kg )( 9.8m/s2)( 0.17 m) = 0.04 J

c) the potential energy when the block is at point P isgiven by

       W = mgh

4 mgR

           =4 (0.012 kg )( 9.8m/s2)( 0.17 m) = 0.08 J

d) the potential energy when the block is at pointQ is given by

         W = mgR

              =  (0.012 kg )( 9.8m/s2)( 0.17 m) = 0.02 J

e)  the potential energy when the block is atthe top of the loop is

       W = mg(2R)

           =2mgR

         =2  (0.012 kg )( 9.8m/s2)( 0.17 m)

= 0.04 J

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