A body of mass 2 kg is moving along x -direction such that its displacement as function of time is given by x(t)= t²+ t+ m , where =1 ~m / s ², =1 ~m / s and =1 ~m . The work done on the body during the time interval t=2 ~s to t=3 ~s , is _ _ _ _ J.

Step-by-step solution

Given x(t) = t^2 + t + 1 , with m = 2 kg, find work from t = 2 s to t = 3 s. Velocity: v(t) = dx dt = 2t + 1 At t = 2 s: v_i = 2(2) + 1 = 5 m/s At t = 3 s: v_f = 2(3) + 1 = 7 m/s Using the work-energy theorem: W = KE = 1 2 m(v_f^2 - v_i^2) = 1 2 (2)(7^2 - 5^2) = 1(49 - 25) = 24 J