Question

force act on 30 gram particle in such a way that the position of particle as a function of time is given by X equals to 30 square minus 40 square + PQ where X is in metre and I is in second the work done using the first four seconds​

Answer 1

 

M = 20 gm = 0.020 kg

x = 3 t - 4 t² + t³   m

v = dx/dt = 3 - 8 t + 3 t²   m/s

a = dv/dt = -8 + 6 t   m/s²

Force = m a = 0.120 t - 0.160  N

Work done = dW = F dx = F * (dx/dt) * dt

        dW = (0.120 t - 0.160) * (3 - 8 t + 3 t²) dt

               = 0.360 t³ -  1.440 t² + 1.640 t - 0.480

W = integral of dW from t = 0 to 4 sec:

     = 0.090 t⁴ - 0.490 t³ + 0.82 t² - 0.480 t ] ,  t = 0 to 4s

     = 2.88 J