Question

A force is applied to a 3.5 kg object moving in the x-direction with the following profile over time: a. What value of F_max gives an impulse of 6.0 N s? b. If the object had an initial speed of 4.2 m/s, what was the change in kinetic energy after the force was applied?

Answer 1

a) the impluse = F_max times t 6 = F_max times 2 F_max = 6/2 = 3N Therefore F_max = 3N b) u = 4.2m/s, m = 3.5 kg, t = 2, F_max = 3N F_max = ma F_max = m(v - u/t) 3 = 3.5 (v - 42/2) v - 42 = 3 times 2/3.5 = 1.714 v = 1.714 + 4.2 v = 5.9 m/s change in kinetic energy is delta KE = kE_f - kE_i = 1/2 mv_f^2 - 1/2 m v_i^2 = 1/2 times m[v_f^2 - v_i^2] = 3.5/2 [(5.9)^2 - (4.2)^2] = 3.5/2[34.81 - 17.64] = 3.5 times 17.17/2 = 30.04 Joule