A particle of mass 2 kg moving along the positive x-axis has an initial velocity of 6 m/s at x = 0 . It is subjected to a position-dependent force F . The force F varies with position x such that the F-x graph forms a triangle: the force is zero at x = 0 , reaches a minimum value of -10 N at x = 2 m , and returns to zero at x = 4 m . The velocity of the particle at x = 4 m is

Step-by-step solution

The work done by the variable force is given by the area under the F-x graph. Since the graph is a triangle with base from x = 0 to x = 4 m and a peak of -10 N , the area is: W = 1 2 base height = 1 2 4 (-10) = -20 J According to the work-energy theorem, the net work done is equal to the change in kinetic energy: W = K_f - K_i -20 = 1 2 m v_f^2 - 1 2 m v_i^2 Given m = 2 kg and v_i = 6 m/s : -20 = 1 2 (2) v_f^2 - 1 2 (2) (6)^2 -20 = v_f^2 - 36 v_f^2 = 16 v_f = 4 m/s Answer: 4 m/s