Question

A small 0.40-kg box is launched from rest by a horizontal spring as shown in the figure below. The block slides on a track down a hill and comes to rest at a distance d from the base of the hill. The coefficient of kinetic friction between the box and the track is 0.37 along the entire track. The spring has a spring constant of 35.0 N/m and is compressed 30.0 cm with the box attached. The block remains on the track at all times. 30.0 cm 50.0 cm 40° (a) What would you include in the system? Explain your choice This answer has not been graded yet. (b) Calculate d. (c) Compare your answer with a similar situation where the inclined part of the track is frictionless

Answer 1

Value of acceleration due to gravity should be mentioned velocity at B 1/2 m v_B^2 = 1/2 kx^2 - mu mg x u_b^2 = kx^2/m - 2 mu g_x = 35 times (3)^2/0.4 - 2 times .37 times 9.8 times .3 u_B^2 = 5.6994 U_B = 2.387 m/s acceleration along BC a = g (sin 40 - mu cos 40 degree) = 9.8 (.64 - .37 times 0.766) = 9.8 times .356 = 3.49 m/s^2 In case of frictionless inclined plane acc. Along the plane = 9.8 times .64 = 6.272 m/s^2 velocity at C V_c^2 = U_B^2 + 2as = 5.6994 + 2 times 6.272 times .5 u_c^2 = 11.9714 From C rightarrow E U_E^2 - V_C^2 = 2 times ad 0 - 11.9714 = .2 times (.37 times 9.8) times d d times = 11.9714/7.252 = 1.65 m = 165 cm Velocity at C u_c^2 - u_b^2 = 2as u_c^2 = u_b^2 + 2as = 5.6994 + 2