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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 tight 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?
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a tight 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?
CBSE | Class 11 | Laws of Motion | NCERT | Physics
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a tight 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?

Given,

A body mass of 10 kilogrames (m1)

B, m2 = 20 kg,

600 N horizontal force

m = m1 + m2 = 30 kg is the total mass of the system.

Using Newton’s second rule of motion, we can calculate

ma = f

600/30 = 20 m/s2 a = F/m

When the force is applied to A, I (10 kg)

m1a = F – T

m1a = T = F

T = 600+

= 600 – 200 = 400 N

(ii) When the force is applied on B (20 kg)

F – T = m2a

T = F – m2a

= 600 – (20 x 20) = 200 N

i

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