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Karnataka Board PUCPUC Science Class 11

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

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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

  1. A,
  2. B along the direction of string. What is the tension in the string in each case?
Numerical
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Solution

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 "m/s"^2`

(i) When force F is applied to body A:

The equation of motion can be written as:

F-T = m1a

T = F - m1a

= 600 – 10 × 20 = 400 N

(ii) When force F is applied to body B:

The equation of motion can be written as:

F – T = m2a

T = F – m2a

∴T = 600 – 20 × 20 = 200 N

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Chapter 4: Laws of Motion - EXERCISES [Page 69]

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NCERT Physics Part 1 and 2 [English] Class 11
Chapter 4 Laws of Motion
EXERCISES | Q 4.15 | Page 69

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