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(D) The work done by a variable force is equal to the area under the force-displacement $(F-x)$ graph.Work done $W = \int F \, dx = 	ext{Area under t

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$5$

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(D) The work done by a variable force is equal to the area under the force-displacement $(F-x)$ graph.Work done $W = \int F \, dx = 	ext{Area under the graph}$.Areas above the $x$-axis are positive,and areas below the $x$-axis are negative.$1$. Area from $x=0$ to $x=1$: Triangle with base $1$ and height $10$. Area $A_1 = \frac{1}{2} 	imes 1 	imes 10 = 5\,J$.$2$. Area from $x=1$ to $x=2$: Rectangle with width $1$ and height $10$. Area $A_2 = 1 	imes 10 = 10\,J$.$3$. Area from $x=2$ to $x=3$: Triangle with base $1$ and height $10$. Area $A_3 = \frac{1}{2} 	imes 1 	imes 10 = 5\,J$.$4$. Area from $x=3$ to $x=4$: Triangle with base $1$ and height $-10$. Area $A_4 = \frac{1}{2} 	imes 1 	imes (-10) = -5\,J$.$5$. Area from $x=4$ to $x=5$: Rectangle with width $1$ and height $-10$. Area $A_5 = 1 	imes (-10) = -10\,J$.Total Work $W = A_1 + A_2 + A_3 + A_4 + A_5 = 5 + 10 + 5 - 5 - 10 = 5\,J$.

$A$ position-dependent force $F$ acts on a particle,and its force-position curve is shown in the figure. The work done on the particle for a displacement from $0$ to $5\,m$ is ............. $J$.

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