Figure gives the acceleration of a 2.0 kg body as it moves from rest along x− axis, while a variable force acts on it from x=0m to x=9m.
The work done by the force on the body when it reaches x=4m and x=7m is
A
21J and 33J respectively
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B
21J and 15J respectively
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C
42J and 60J respectively
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D
42J and 30J respectively
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Solution
The correct option is D42J and 30J respectively
Here we have acceleration versus displacement graph.
Work done by the force on the body,
when it reaches at x=4 m is Wt=4 = Mass of the body × Area under ABCD Wt=4=2[(12×1×6)+(3×6)]=42J
Work done by the force on the body when it reaches at x=7m is Wt=7=Wt=4+ (Mass of the body )(Area under CDE - Area under EFGH) =42+2[(12×1×6)−(12×1×6)−(1×6)] Wt=7=30 J