### 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:

*m*

_{1}= 8 kg

Larger mass, *m*_{2} = 12 kg

Tension in the string = *T*

Mass *m*_{2}, owing to its weight, moves downward with acceleration *a*,and mass *m*_{1 }moves upward.

Applying Newton’s second law of motion to the system of each mass:

__For mass m _{1}:__

The equation of motion can be written as:

*T*–

*m*

_{1}g =

*ma*…

**(i)**

__For mass m _{2}:__

The equation of motion can be written as:

*m*

_{2}g

*–*

*T*=

*m*

_{2}

*a*…

**(ii)**

Adding equations (*i*) and (*ii*), we get:

(*m*_{2} – *m*_{1})g = (*m*_{1} + *m*_{2})*a*

∴ *a* = ( (*m*_{2} – *m*_{1}) / (*m*_{1} + *m*_{2}) )g ….**(iii)**

= (12 – 8) / (12 + 8) × 10 = 4 × 10 / 20 = 2 ms^{-2}

Therefore, the acceleration of the masses is 2 m/s^{2}.

Substituting the value of *a* in equation (*ii*), we get:

*m _{2}g* –

*T*=

*m*(

_{2}*m*–

_{2}*m*)g / (

_{1}*m*+

_{1}*m*)

_{2}T = (m

_{2}– (

*m*–

_{2}^{2}*m*) / (

_{1}m_{2}*m*+

_{1}*m*) )g

_{2}= 2

*m*g / (

_{1}m_{2}*m*+

_{1}*m*)

_{2}= 2 × 12 × 8 × 10 / (12 + 8)

= 96 N

Therefore, the tension in the string is 96 N.