22) Two masses are attached by a weightless string over a frictionless pulley as shown. If $\mathrm{m} 1=78 \mathrm{~kg}$ and $\mathrm{m} 2=42 \mathrm{~kg}$ calculate:
a. The acceleration of $\mathrm{m} 2$ if the table is frictionless.
b. The acceleration of $\mathrm{m} 1$ if the table has a coefficient of friction = 0.465 .

Step 1 ：Given that the mass m1 = 78 kg, mass m2 = 42 kg, and the acceleration due to gravity g = 9.8 m/s^2.

Step 2 ：For part a, since the table is frictionless, the only forces acting on the system are the weights of the two masses. The net force acting on the system is the difference in the weights of the two masses, and this net force causes the system to accelerate. We can use Newton's second law, F = ma, to find the acceleration.

Step 3 ：Calculate the net force for part a: F_net_a = m1*g - m2*g = 78*9.8 - 42*9.8 = 352.8 N.

Step 4 ：Calculate the total mass of the system: m_total = m1 + m2 = 78 + 42 = 120 kg.

Step 5 ：Calculate the acceleration of m2: a_m2 = F_net_a / m_total = 352.8 / 120 = 2.94 m/s^2.

Step 6 ：For part b, we need to consider the force of friction acting on m1. The force of friction is given by the product of the coefficient of friction and the normal force. In this case, the normal force is equal to the weight of m1. The net force acting on the system is now the difference in the weights of the two masses minus the force of friction. We can again use Newton's second law to find the acceleration.

Step 7 ：Calculate the force of friction: F_friction = mu * m1 * g = 0.465 * 78 * 9.8 = 355.446 N.

Step 8 ：Calculate the net force for part b: F_net_b = F_net_a - F_friction = 352.8 - 355.446 = -2.646 N.

Step 9 ：Calculate the acceleration of m1: a_m1 = F_net_b / m_total = -2.646 / 120 = -0.022 m/s^2.

Step 10 ：The final answer is: The acceleration of m2 when the table is frictionless is approximately \(\boxed{2.94}\) m/s^2. The acceleration of m1 when the table has a coefficient of friction of 0.465 is approximately \(\boxed{-0.022}\) m/s^2.