The kinetic energy of a 2 kg particle moving along the positive x -axis is given as a function of displacement x by the equation K(x) = 150 - x^4 J . The magnitude of the particle's acceleration at x = 3 m is ________ m/s ^2 .

Step-by-step solution

The kinetic energy is given by K = 1 2 mv^2 . Differentiating kinetic energy with respect to displacement x gives the net force: dK dx = d dx ( 1 2 mv^2 ) = mv dv dx = ma = F Given K(x) = 150 - x^4 , we differentiate it with respect to x : dK dx = -4x^3 Equating this to ma : ma = -4x^3 Substitute m = 2 kg : 2a = -4x^3 a = -2x^3 At x = 3 m , the acceleration is: a = -2(3)^3 = -2(27) = -54 m/s ^2 The magnitude of the acceleration is 54 m/s ^2 . Answer: 54