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
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a
16 J
b
32 J
c
1.6 J
d
8 J
answer is A.
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Detailed Solution
Work done = Area under curve and displacement axis= Area of trapezium= 12×(sum of two parallel lines) × distance between them= 12(10+4)×(2.5−0.5)= 1214×2 = 14 J As the area actually is not trapezium so work done will be more than 14 J i.e. approximately 16 J