A closed circular tube of average radius 15,c | vedclass

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(D) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy.$W_{	ext{total}} = \Delta KE =

Vedclass · Accepted Answer

$-245$

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(D) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy.$W_{	ext{total}} = \Delta KE = KE_f - KE_i$Here,$W_{	ext{total}} = W_{	ext{friction}} + W_{	ext{gravity}}$.The block starts at the top and stops at the bottom,so the net change in height is $h = 2r = 2 	imes 0.15\,m = 0.3\,m$.The work done by gravity is $W_g = mg 	imes h = 1 	imes 10 	imes 0.3 = 3\,J$.The initial kinetic energy is $KE_i = \frac{1}{2}mv^2 = \frac{1}{2} 	imes 1 	imes (22)^2 = \frac{484}{2} = 242\,J$.The final kinetic energy is $KE_f = 0\,J$ (since it stops).Applying the work-energy theorem:$W_f + W_g = KE_f - KE_i$$W_f + 3 = 0 - 242$$W_f = -242 - 3 = -245\,J$.Therefore,the work done by the tube (friction) on the block is $-245\,J$.

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