Question

Two masses m1 and m2 (m1 > m2) are connected to the ends of a light inextensible string which passes over the surface of a smooth fixed pulley. If the system is released from rest, the acceleration of the centre of mass of the system will be (g= acceleration due to gravity)

Answer 1

Answer:

$a=\dfrac{(m_1-m_2)g}{m_1+m_2}\ m/s^2$

Explanation:

Given that

Mass of first block = m₁

Mass of first second = m₂

Lets T is the tension in the spring

m₁ >m₂

It means that mass m₁ will move downward and mass m₂ will move upward.

Lets a is acceleration of the system

For mass m₁:

m₁ g -T=m₁a       ---------1

For mass m₂ :

T - m₂ g= m₂a    ---------2

By adding equation 1 and 2

m₁ g - m₂ g = m₁a + m₂a    

(m₁ -m₂) g = (m₁+m₂) a

$a=\dfrac{(m_1-m_2)g}{m_1+m_2}\ m/s^2$

So the acceleration of the system is $a=\dfrac{(m_1-m_2)g}{m_1+m_2}\ m/s^2$