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(C) The work done on the body is equal to the change in its kinetic energy.According to the work-energy theorem, $W = \Delta K = K_f - K_i$.Since the

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$10 \, m/s$

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(C) The work done on the body is equal to the change in its kinetic energy.According to the work-energy theorem, $W = \Delta K = K_f - K_i$.Since the body starts from rest, the initial kinetic energy $K_i = 0$.The work done is equal to the area under the force-displacement $(F-x)$ graph.The area under the graph is a trapezoid with parallel sides of length $3 \, m$ (from $x=3$ to $x=6$) and $9 \, m$ (from $x=0$ to $x=9$), and height $20 \, N$.Area $= \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$Area $= \frac{1}{2} 	imes (3 + 9) 	imes 20 = \frac{1}{2} 	imes 12 	imes 20 = 120 \, J$.Thus, the work done $W = 120 \, J$.Equating work done to kinetic energy: $120 = \frac{1}{2} m v^2$.Given $m = 2.4 \, kg$, we have $120 = \frac{1}{2} 	imes 2.4 	imes v^2$.$120 = 1.2 	imes v^2$.$v^2 = \frac{120}{1.2} = 100$.$v = 10 \, m/s$.

$A$ body of mass $2.4 \, kg$ is subjected to a force which varies with distance as shown in the figure. The body starts from rest at $x=0$. Its velocity at $x=9 \, m$ is

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