Q. 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
A) $16\ J$
B) $32\ J$
C) $1.6\ J$
D) $8\ J$
Explanation
(a) Work done = Area under curve and displacement axis
= Area of trapezium
$= \frac{1}{2} \times (sum\ of\ two\ parallel\ lines) \times distance\ between\ them$
$= \frac{1}{2}\ (10 + 4) \times (2.5 - 0.5) = \frac{1}{2}14 \times 2 = 14\ J$
As the area actually is not trapezium so work done will be more than $14\ J$ $i.e.$ approximately $16\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.