Question

Question 3. [12 marks] in the diagram, two blocks (m - 4 kg and m 10 kg) are a an inextensible cord of negligible mass that passes Block m, sits on a horizontal, rough surface with μ.-0.300 and :0.5 to an ideal spring (k 40 N/m) that is initially stretched relative initial tension in the spring 16 N. The system is then released from rest. Deternm ttached by over an ideal (massless, frictionless) pulley to its resting length, so that the of the blocks when m2 has fallen through a vertical distance of 50 cm. n2

Answer 1

m_1 = 4 kg m_2 = 10 kg mu_k = 0.3 mu_s = 0.5 k = 40 N/m initial tension in the spring = kx 16 = 40 middot x_0 x_0 = 16/40 = 0.4 m x_0 = initial extension in the spring = 0.4 m x = 0.5 m From work energy theorem Work done by all forces = Delta K.E = change in kinetic energy w (N_1) + w (F_s) + W (f_k) + w (m_1 g) + w (m_2 g) = Delta K.E = 1/2 (m_2) V^2 + 1/2 m_1 V^2 + (0 + 0) (1) w (N_1) = work done by normal N_1 = 0 w (m_1 g) = work done by m_1 g = 0 w (F_s) = work done by spring force = -Delta u = -1/2 - (1/2 k (x + x_0)^2 - 1/2 x x_0^2) w (f_k) = work done by friction force = -mu_k m_1 g middot x