The figure shows the F-x graph,where F is the | vedclass

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(A) The work done by a variable force is equal to the area under the $F-x$ graph.Work done $(W)$ = Area of the rectangle from $x=1$ to $x=2$ + Area of

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$10$

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(A) The work done by a variable force is equal to the area under the $F-x$ graph.Work done $(W)$ = Area of the rectangle from $x=1$ to $x=2$ + Area of the rectangle from $x=2$ to $x=3$ + Area of the rectangle from $x=3$ to $x=4$.Area $1$ ($x=1$ to $2$): $F = 10 \; N$,$\Delta x = 1 \; m$. Area = $10 	imes 1 = 10 \; J$.Area $2$ ($x=2$ to $3$): $F = -10 \; N$,$\Delta x = 1 \; m$. Area = $-10 	imes 1 = -10 \; J$.Area $3$ ($x=3$ to $4$): $F = 10 \; N$,$\Delta x = 1 \; m$. Area = $10 	imes 1 = 10 \; J$.Total Work Done = $10 + (-10) + 10 = 10 \; J$.

The figure shows the $F-x$ graph,where $F$ is the force applied and $x$ is the distance covered by the body along a straight-line path. Given that $F$ is in newton $(N)$ and $x$ is in metre $(m)$,calculate the total work done in joules $(J)$.

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