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? | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-5 Laws Of Motion



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 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 = 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 = m1a
T = Fm1a
= 600 – 10 × 20 = 400 N … (i)


When force F is applied on body B:

The equation of motion can be written as:

FT = m2a
T = Fm2a
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.