InterviewSolution
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InterviewSolution
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The displacement of a particle of mass `1kg` on a horizontal smooth surface is a function of time given by `x=1/3t^3`. Find out the work done by the external agent for the first one second. |
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Answer» Given that `x=1/3t^3`. The velocity of the particle at any instant t is `v=(dx)/(dt)=t^2` The acceleration of the particle at any instant t is `a=(d^2x)/(dt^2)=2t` Therefore, work done by the force imposed is `W=intFdx=intF(dx)/(dt)*dt=intm(a)((dx)/(dt))(dt)` Putting the values of `m=1kg`, `(dx)/(dt)`, and a, we obtain `W=underset0overset(t=1)int(1)(t)^2(2t)dt=2underset0overset1intt^3dt=2(t^4/4)_0^1=0.5J` |
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