#### Question

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?

#### Solution 1

Horizontal force, *F* = 600 N

Mass 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 "m/s"^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)**

#### Solution 2

Accleration = `(600 N)/(10 kg + 20 kg) = 20 ms^(-2)`

(i) When force is applied on 10 kg mass

600 - T = 10 x 20 or

T = 400 N

(ii)When force is applied on 50 kg mass

600 - T = 20 x 20 or

T = 200 N