InterviewSolution
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InterviewSolution
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A body of mass 0.5 kg travels in a straight line with velocity `v= kx^(3//2)` where `k=5m^(-1//2)s^(-1)`. The work done by the net force during its displacement from x=0 to x=2 m isA. 1.5 JB. 50 JC. 10 JD. 100 J |
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Answer» Correct Answer - B Given :m=0.5 kg, `v=kx^(3//2)` where , k=`5 m^(-1//2) s^(-1)` Acceleration , `a=(dv)/(dt)=(dv)/(dx)(dx)/(dt)=v(dv)/(dx) " " (because v=(dx)/(dt))` As `v^2=k^2 x^3` Differentiating both sides with respect to x, we get `2v(dv)/(dx)=3k^2x^2` `therefore` Acceleration, `a=3/2k^2x^2` Force, F=Mass x Acceleration =`3/2mk^2x^2` Work done, `W=int Fdx=underset(0)overset(2)int 3/2 mk^2x^2dx` `W=3/2mk^2[x^3/3]_0^2=3/6xx0.5xx5^2xx[2^3-0]=50 J` |
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