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Under the action of constant force, a 2kg block moves such that its position x as a function of time is given by X=t^3/3 ,where x in meters and t in sec , the workdone by force in first 2 seconds isA 26.6 jB 3.3 jC 16 jD 32 jSolution) v=dx/dt=t^2,a=dv/dt=2t,f(t)=ma(t),w=∫f(t)*v(t) dt (from x=0 to x=2 sec)
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