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Question

Physics Question on Work-energy theorem

Power supplied to a particle of mass 2 kg varies with time as P=3t22W.P=\frac{3{{t}^{2}}}{2}W. Here t is in second. If velocity of particle at t=0t=0 is v=0,v=0, the velocity of particle at time t=2st=2\,s will be

A

1 m/s

B

4 m/s

C

2 m/s

D

22m/s2\sqrt{2}\,m/s

Answer

2 m/s

Explanation

Solution

From work energy theorem
ΔKE=WEnet\Delta KE =W{{E}_{net}}
or KfKi=pdt{{K}_{f}}-{{K}_{i}}=\int_{{}}^{{}}{p\,dt}
12mv20=02(32t2)dt\frac{1}{2}m{{v}^{2}}-0=\int_{0}^{2}{\left( \frac{3}{2}{{t}^{2}} \right)}\,dt
v2=[t32]02{{v}^{2}}=\left[ \frac{{{t}^{3}}}{2} \right]_{0}^{2}
v=2m/sv=2m/s