A particle of mass 2 kg is moving in the xy -plane such that its velocity vector as a function of time t is given by v (t) = (p t^2) i + (q t) j . The work done by all the forces acting on the particle during the time interval from t = 0 to t = 2 s is (Take p = 1 m/s ^3 and q = 2 m/s ^2 ).

Step-by-step solution

According to the work-energy theorem, the work done by all forces on a particle equals its change in kinetic energy. The velocity vector of the particle is given by: v (t) = (1 t^2) i + (2 t) j = t^2 i + 2t j At t = 0 , the initial velocity is: v (0) = 0 i + 0 j = 0 Initial kinetic energy, K_i = 0 At t = 2 s , the final velocity is: v (2) = (2^2) i + (2 2) j = 4 i + 4 j The square of the final speed is: | v (2)|^2 = 4^2 + 4^2 = 16 + 16 = 32 (m/s) ^2 Final kinetic energy, K_f = 1 2 m| v (2)|^2 = 1 2 2 32 = 32 J Work