Questions
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
detailed solution
Correct option is D
At t=1sec, V1=5×1 m/s=5 m/sAt t=2sec, V2=5×2 m/s=10 m/sBy work - energy theoremWork done =12mV22−12mV12=12×0.5×102−52J⇒W=18.75 JTalk to our academic expert!
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