Question 5.15:

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?


Horizontal force, F = 600 N

Mass of body A, m1 = 10 kg

Mass of body B, m2 = 20 kg

Total mass of the system, m = m1+m2 = 30 kg

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

 F = ma
a = Fm = 60030 = 20 m/s2

(a) When force F is applied on body A:

The equation of motion can be written as:

F – T = m1a

Or, T = F – m1a

= 800 – 20 x 10 = 400 N

(b) When force F is applied on body B:

The equation of motion can be written as:

F – T = m2a

T = F – m2a

T = 600 – 20 × 20 = 200 N.