﻿ 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? from Physics Laws of Motion Class 11 Manipur Board

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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,

F = ma

Acceleration (
a) produced in the system can be calculated as,

=

When force is applied on a body A.

Equation of motion can be written as,

F - T = m1a

Therefore,

T = F - m1a

= 600 - 10 x 20

= 400 N                                ...  (i)

When Force is applied on a body B, we have

F - T = m2a

i.e., T = F - m2

Therefore,

T = 600 - 20 x 20 = 200 N         ... (ii)

From (i) and (ii), we can say that the answer is different in both the cases.

Therefore, the answer depends on which end of mass, the force is applied.
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