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Q.

A particle of mass 0.5 kg is moving along x-axis and at time t, its velocity is V = 5 t m/s where t is time in second. Then work done by the applied force in the time interval t = 1 sec to t = 2 sec is

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a

20.5 J

b

25 J

c

6.25 J

d

18.75 J

answer is D.

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Detailed Solution

At t=1sec, V1=5×1 m/s=5 m/s

At t=2sec, V2=5×2 m/s=10 m/s

By work - energy theorem

Work done =12mV22−12mV12=12×0.5×102−52J

⇒​W=18.75 J

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