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(A) The work done by a variable force is equal to the area under the force-displacement graph.For the interval from $x = 1 \; cm$ to $x = 5 \; cm$:$1$

Vedclass · Accepted Answer

$20$

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(A) The work done by a variable force is equal to the area under the force-displacement graph.For the interval from $x = 1 \; cm$ to $x = 5 \; cm$:$1$. From $x = 1$ to $x = 2$,the area is a rectangle with height $10 \; dyne$ and width $1 \; cm$: $Area_1 = 10 	imes 1 = 10 \; erg$.$2$. From $x = 2$ to $x = 3$,the area is a rectangle with height $20 \; dyne$ and width $1 \; cm$: $Area_2 = 20 	imes 1 = 20 \; erg$.$3$. From $x = 3$ to $x = 4$,the area is a rectangle below the x-axis with height $-20 \; dyne$ and width $1 \; cm$: $Area_3 = -20 	imes 1 = -20 \; erg$.$4$. From $x = 4$ to $x = 5$,the area is a rectangle with height $10 \; dyne$ and width $1 \; cm$: $Area_4 = 10 	imes 1 = 10 \; erg$.Total work done $W = Area_1 + Area_2 + Area_3 + Area_4 = 10 + 20 - 20 + 10 = 20 \; erg$.

The figure shows the relationship between force and position. What is the work done by the force for the displacement of the object from $x = 1 \; cm$ to $x = 5 \; cm$ in ergs?

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