Work-power-energy, Mechanical Engineering

 

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 is

A 26.6 j

B 3.3 j

C 16 j

D 32 j

Solution) 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)

Posted Date: 3/11/2013 1:55:51 AM | Location : United States







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