A particle of mass 2 kg moves along the positive x -axis under the action of a position-dependent force F . The force increases linearly from 0 to 10 N between x = 0 and x = 2 m , remains constant at 10 N from x = 2 m to x = 4 m , and then remains constant at -5 N from x = 4 m to x = 8 m . If the initial velocity of the particle at x = 0 is 4 m/s , the kinetic energy of the particle at x = 8 m is

Step-by-step solution

According to the work-energy theorem, the net work done by the force equals the change in kinetic energy of the particle: W_ net = K_f - K_i . The initial kinetic energy is: K_i = 1 2 m v^2 = 1 2 2 (4)^2 = 16 J The work done is the algebraic sum of the areas under the force-position graph: Work from x = 0 to x = 2 m (area of triangle) = 1 2 2 10 = 10 J Work from x = 2 to x = 4 m (area of rectangle) = 2 10 = 20 J Work from x = 4 to x = 8 m (area of rectangle) = 4 (-5) = -20 J Total work done W_ net = 10 + 20 - 20 =