### Q.15:- Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

**Answer:-**Horizontal force,

*F*= 600 NMass of body A, *m*_{1} = 10 kg

Mass of body B, *m*_{2} = 20 kg

Total mass of the system, *m* = *m*_{1} + *m*_{2} = 30 kg

Using Newton’s second law of motion, the acceleration (*a*) produced in the system can be calculated as:

*F* = *ma*

∴ *a* = *F* / *m* = 600 / 30 = 20 ms^{-2}

When force *F* is applied on body A:

The equation of motion can be written as:

*F – T *= *m*_{1}*a*

∴*T* = *F* – *m*_{1}*a*

= 600 – 10 × 20 = 400 N … **(i)**

When force *F* is applied on body B:

The equation of motion can be written as:

*F* – *T* = *m*_{2}*a*

*T* = *F* – *m*_{2}*a*

∴*T* = 600 – 20 × 20 = 200 N … **(ii)**

which is different from value of *T* in case** (i)**. Hence our answer depends on which mass end, the force is applied.