Question

223. The 0.60 kg collar is attached by a spring to point B. The collar starts at rest at A, and then siides down the shaft. The average force of fricticn is (F is 0.80 Newtons Determine the velocity of the collar at C 0.84m Spring Constand- 40 N/m Unstretched length of spring 0.63m 0.58m V 175/s

Answer 1

{friction f = 0.8 N, m = 0.6 kg} {L_0 = unstretched length = 0.65 m} energy stored in spring when collar is at A, v_1 = 1/2 k[x_2 - L_0]^2 v_1 = 1/2 times 40 times [0.84 - 0.65]^2 V_1 = 0.722 J energy stored in spring when collar is at ?B, v_2 = 1/2 K [x_1 - L_0]^2 v_2 = 1/2 times 40 times [1.021 - 0.65]^2 where x^2_1 = H^2 + x^2_2 x_1 = squareroot 0.58^2 + 0.84^2 x_1 = 1.021 m v_2 = 2.7496.3 work done by spring force, w_s = v_1 - v_2 = > w_s = 0.722 - 2.7496 = > w_s = -2.0276 work done by friction from A to B W_f = -fH = > w_f = -0.8 times 0.85 = > w_f = -0.464 J work done by gravits, w_g = mgH = > = 0.6 times 9.81 times 0.58 = > w_g = 3.41388 J \r\n