Question

A load of bricks with mass m1 = 14.0 kg hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass m2= 28.4 kg is suspended from the other end of the rope, as shown in the figure. The system is released from rest. Use g = 9.80 m/s2 for the magnitude of the acceleration due to gravity.

Answer 1

Mass m1 and m2 are connected together and the rope is passing over the pulley

here we can see that the two masses are connected so the acceleration will be same but opposite in direction

So here we can use the force equations to solve for the acceleration of two blocks

now for block of mass m1

$T - m_1g = m_1a$

for block of mass m2

$m_2g - T = m_2a$

Now add above two equations to find its acceleration

$T - m_1g + m_2g - T = m_1a + m_2a$

$(m_2 - m_1)*g = (m_1 + m_2)*a$

$a = \frac{(m_2 - m_1)*g}{(m_1 + m_2)}$

now we will plug in all the values

$a = \frac{(28.4 - 14)*9.8}{28.4 + 14)}$

$a= 3.33 m/s^2$

So the acceleration of two blocks will be 3.33 m/s^2