Question

In the modified atwood machine, the weight of mass A is 288 grams. The weight of mass B is 124 grams, and it is hanging 58 cm above the ground. When the system is released from rest, it takes 1.75 seconds for mass B to hit the floor. Determine the coeffecient of kinetic friciton between mass A and the table, and determine the energy dissipated by the system.

Answer 1

a be the acceleration m_B g - T = m_B a - (1) and, T - muN = m_A a & N = m_A g or, T = m_A (a + mug) - (2) solving for a using (1) & (2) m_B g - m_A a - m_A mu g = m_B a or, (m_A + m_B) a = g (m_B - m_B mu) or, a = g(m_B - m_A mu)/m_A + m_B Now, s = 1/2 at^2 or a = 2s/t^2 = 2 times 0.58 m/1.75^2 s^2 = 0.3788 m/s^2 or g(m_B - m_A mu)/m_A + m_B = 2s/t^2 or m_B - m_A mu = 2s/t^2 m_A + m_B/g or mu = 1/m_A [m_B - (2s/t^2 . m_A + m_B/g] = 1/0.288 [0.124 - { 0.3788 times { 0.3788 times (0.288 + 0.124)/7.81}]