Question

in terms of m1 , m2 , and g , find the acceleration of the first block in the figure (figure 1). there is no friction anywhere in the system.

Answer 1

The acceleration of the first block in terms of the first block and the gravity g is
`a = (m_2g - m_1g)/(m_1 + m_2)`

How to determine the acceleration of the first block

From the question, we have the following parameters that can be used in our computation:

The figure

Where, we have the frictionless pulley with a negligble mass

The direction of the acceleration is towards the mass m2 i.e. downward

So, we have the following equations

`T = m_1a + m_1g`

`m_2g = T + m_2a`

Where

T is the tension in the pulley

By substitution, we have

`m_2g = m_1a + m_1g + m_2a`

This gives

`m_1a + m_2a = m_2g - m_1g`

Factor out a from the left hand side

`a(m_1 + m_2) = m_2g - m_1g`

So, we have

`a = (m_2g - m_1g)/(m_1 + m_2)`

Hence, the acceleration of the first block is
`a = (m_2g - m_1g)/(m_1 + m_2)`