The relation between the displacement X of an object produced by the application of the variable force F is represented by a 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
Text Solution
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Work done = Area under curve and displacement axis
= Area of trapezium
= $\frac{1}{2}$ × (sum of two parallel lines) × distance between them
= $\frac{1}{2}(10+4) \times (2.5 - 0.5)$
= $\frac{1}{2} 14 \times 2$ = 14 J
As the area actually is not trapezium so work done will be more than 14 J i.e. approximately 16 J
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