A body of mass 2 kg moves along the x -axis such that its displacement as a function of time is given by x(t) = t^3 + 18t (where x is in meters and t is in seconds). If the total work done on the body during the time interval from t = 0 to t = 2 s is 576 J, the positive value of the constant is

Step-by-step solution

The velocity of the body is given by the derivative of displacement with respect to time: v(t) = dx dt = 3 t^2 + 18 At t = 0 s, the initial velocity is: v_i = 3 (0)^2 + 18 = 18 m/s At t = 2 s, the final velocity is: v_f = 3 (2)^2 + 18 = 12 + 18 m/s According to the work-energy theorem, the work done on the body is equal to its change in kinetic energy: W = KE = 1 2 m(v_f^2 - v_i^2) Substitute the given values into the equation: 576 = 1 2 (2)[(12 + 18)^2 - 18^2] 576 = (12 + 18)^2 - 324 (12 + 18)^2 = 576 + 324 = 900