A 10 , kg mass moves along the x-axis. Its ac | vedclass

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Question

(A) According to the work-energy theorem,the work done by the net force is equal to the change in kinetic energy. Alternatively,work done $W = \int F

Vedclass · Accepted Answer

$8 	imes 10^{-2} \, J$

Vedclass · Answer

(A) According to the work-energy theorem,the work done by the net force is equal to the change in kinetic energy. Alternatively,work done $W = \int F \, dx = \int m a \, dx = m \int a \, dx$.Here,$m = 10 \, kg$. The integral $\int a \, dx$ represents the area under the acceleration-displacement graph.The graph is a triangle with base $b = 8 \, cm = 8 	imes 10^{-2} \, m$ and height $h = 20 \, cm/s^2 = 0.2 \, m/s^2$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (8 	imes 10^{-2} \, m) 	imes (0.2 \, m/s^2) = 0.8 	imes 10^{-2} \, m^2/s^2$.Work done $W = m 	imes 	ext{Area} = 10 \, kg 	imes (0.8 	imes 10^{-2} \, m^2/s^2) = 8 	imes 10^{-2} \, J$.

$A$ $10 \, kg$ mass moves along the $x-$axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from $x = 0$ to $x = 8 \, cm$?

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