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(A) The work done by a variable force is equal to the area under the force-displacement $(F-X)$ graph.To find the work done between $X = 0.5 \, m$ and

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$16$

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(A) The work done by a variable force is equal to the area under the force-displacement $(F-X)$ graph.To find the work done between $X = 0.5 \, m$ and $X = 2.5 \, m$,we calculate the area under the curve between these two points.At $X = 0.5 \, m$,the force $F \approx 10 \, N$.At $X = 2.5 \, m$,the force $F \approx 4 \, N$.Approximating the area under the curve as a trapezium:Area $\approx \frac{1}{2} 	imes (F_1 + F_2) 	imes (X_2 - X_1)$Area $\approx \frac{1}{2} 	imes (10 + 4) 	imes (2.5 - 0.5)$Area $\approx \frac{1}{2} 	imes 14 	imes 2 = 14 \, J$.Since the curve is concave upwards,the actual area under the curve is slightly larger than the area of the trapezium formed by the straight line connecting the two points.Therefore,the work done is approximately $16 \, J$.

The relation between the displacement $X$ of an object produced by the application of the variable force $F$ is represented by the graph shown in the figure. If the object undergoes a displacement from $X = 0.5 \, m$ to $X = 2.5 \, m$,the work done will be approximately equal to .............. $J$.

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