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(A) The work done by a variable force is equal to the area under the $F-x$ graph,considering the algebraic sign of the area.From the graph,the area fr

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(A) The work done by a variable force is equal to the area under the $F-x$ graph,considering the algebraic sign of the area.From the graph,the area from $x = 0$ to $x = 4\,m$ is a triangle above the $x$-axis:Area$_1 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4\,m 	imes 20\,N = 40\,J$.The area from $x = 4\,m$ to $x = 8\,m$ is a triangle below the $x$-axis:Area$_2 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (8 - 4)\,m 	imes (-20\,N) = \frac{1}{2} 	imes 4\,m 	imes (-20\,N) = -40\,J$.Total work done $W = 	ext{Area}_1 + 	ext{Area}_2 = 40\,J + (-40\,J) = 0\,J$.

$A$ force $F$ acting on an object varies with distance $x$ as shown in the figure. The work done by the force in moving the object from $x = 0$ to $x = 8\,m$ is ......... $J$.

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