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(C) According to the work-energy theorem,the total work done on the body is equal to the change in its kinetic energy.$W_{	ext{gravity}} + W_{	ext{a

Vedclass · Accepted Answer

$-1500$

Vedclass · Answer

(C) According to the work-energy theorem,the total work done on the body is equal to the change in its kinetic energy.$W_{	ext{gravity}} + W_{	ext{air}} = \Delta KE$Given:Mass $m = 10\, kg$,Height $h = 20\, m$,Final velocity $v = 10\, m/s$,Initial velocity $u = 0\, m/s$,$g = 10\, m/s^2$.Work done by gravity $W_{	ext{gravity}} = mgh = 10 	imes 10 	imes 20 = 2000\, J$.Change in kinetic energy $\Delta KE = \frac{1}{2}mv^2 - 0 = \frac{1}{2} 	imes 10 	imes (10)^2 = 500\, J$.Substituting these values into the work-energy theorem:$2000 + W_{	ext{air}} = 500$$W_{	ext{air}} = 500 - 2000 = -1500\, J$.Thus,the work done by the air resistance is $-1500\, J$.

$A$ body of mass $10\, kg$ is released from a tower of height $20\, m$ and the body acquires a velocity of $10\, m/s$ after falling through the distance $20\, m$. The work done by the air resistance on the body is: ................. $J$ (Take $g = 10\, m/s^2$)

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