Question

T. T, rA

find acceleration of the box weighting 20kg.

Answer 1

Given that coefficient of friction mu = 0.3, Mass of block (1) and (2) M_1 = 10 kg M_2 = 20 kg, Block (1) is Moving from left to right, Now Free body diagram of Block (1) Horizontal Force on it T_1 - mu M_1 g = m_1a equation (1) \r\n Now Free body diagram of Block (2) vertical Force acting on Block (2) (M_2 g - T_1) = M_2 a equation (2) By dividing equation (1) and (2) We get T_1 - mu m_1 g/m_2 g - T_1 = m_1 a/m_2 a (T_1 - mu m_1 g)m_2 = (m_2 g - T_1)m_1 (T_1 - 0.3 times 10 times 9.81)20 (20 times 9.81 - T_1)10 20T_1 - 588.6 = 1962 - 10 T_1 30 T_1 = 2550.6 \r\n T_1 = 85.02 N Put T_1 value in equation (2) We get (M_2 g - T_1) = m_2 a 20 times 9.81 - 85.02 = 20 times a a = 196.2 - 85.02/20 = 5.559 m/sec^2 acceleration of the Box 20 kg is 5.559 m/sec^2