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)