A body of mass 2 kg moves along the x -axis such that its position varies with time t according to the equation x(t) = t^3 - 2t^2 + 3t , where x is in meters and t is in seconds. The net work done by the force acting on the body during the time interval from t = 0 to t = 2 s is

Step-by-step solution

The position of the body is given by x(t) = t^3 - 2t^2 + 3t . The velocity of the body as a function of time is the derivative of position with respect to time: v(t) = dx dt = 3t^2 - 4t + 3 The initial velocity at t = 0 is: v(0) = 3(0)^2 - 4(0) + 3 = 3 m/s The final velocity at t = 2 s is: v(2) = 3(2)^2 - 4(2) + 3 = 12 - 8 + 3 = 7 m/s According to the work-energy theorem, the net work done is equal to the change in kinetic energy: W = K = 1 2 m(v_f^2 - v_i^2) Substituting the values: W = 1 2 (2)(7^2 - 3^2) W = 49 -