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(C) The work done by a variable force is equal to the area under the force-displacement graph.To find the work done from $x = 0$ to $x = 35\,m$,we cal

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$287.5$

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(C) The work done by a variable force is equal to the area under the force-displacement graph.To find the work done from $x = 0$ to $x = 35\,m$,we calculate the area of the trapezoid formed by the graph from $x = 0$ to $x = 10$,the rectangle from $x = 10$ to $x = 30$,and the trapezoid from $x = 30$ to $x = 35$.$1$. Area from $x = 0$ to $x = 10$: This is a triangle with base $10\,m$ and height $10\,N$. Area $= \frac{1}{2} 	imes 10 	imes 10 = 50\,J$.$2$. Area from $x = 10$ to $x = 30$: This is a rectangle with width $(30 - 10) = 20\,m$ and height $10\,N$. Area $= 20 	imes 10 = 200\,J$.$3$. Area from $x = 30$ to $x = 35$: This is a trapezoid with parallel sides $10\,N$ and $5\,N$,and height $(35 - 30) = 5\,m$. Area $= \frac{1}{2} 	imes (10 + 5) 	imes 5 = \frac{1}{2} 	imes 15 	imes 5 = 37.5\,J$.Total Work Done $= 50 + 200 + 37.5 = 287.5\,J$.

The adjacent figure shows the force-displacement graph of a moving body. The work done in displacing the body from $x = 0$ to $x = 35\,m$ is equal to ........... $J$.

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