Q.16:- Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Answer:-

The given system of two masses and a pulley can be represented as shown in the following figure:

Larger mass, m₂ = 12 kg
Tension in the string = T
Mass m₂, owing to its weight, moves downward with acceleration a,and mass m₁ moves upward.
Applying Newton’s second law of motion to the system of each mass:

For mass m₁:
The equation of motion can be written as:
T – m₁g = ma … (i)

For mass m₂:
The equation of motion can be written as:
m₂g – T = m₂a … (ii)

Adding equations (i) and (ii), we get:
(m₂ – m₁)g = (m₁ + m₂)a
∴ a = ( (m₂ – m₁) / (m₁ + m₂) )g    ….(iii)
= (12 – 8) / (12 + 8) × 10  =  4 × 10 / 20  =  2 ms^-2
Therefore, the acceleration of the masses is 2 m/s².
Substituting the value of a in equation (ii), we get:
m₂g – T = m₂(m₂ – m₁)g / (m₁ + m₂)
T = (m₂ – (m₂² – m₁m₂) / (m₁ + m₂) )g
= 2m₁m₂g / (m₁ + m₂)
= 2 × 12 × 8 × 10 / (12 + 8)
= 96 N
Therefore, the tension in the string is 96 N.