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?

Answer 1

Horizontal force, \(F \) = \(600 \;N\) 
Mass of body A, \(m_1\) = \(10 \;kg\) 
Mass of body B, \(m_2\) = \(20 \;kg\) 
Total mass of the system, \(m\) = \(m_1 + m_2 \) = \(30 \;kg\) 
Using Newton’s second law of motion, the acceleration (a) produced in the system can be calculated as: 
\(F\) = \(ma\)
\(\therefore\)  \(a\) = \(\frac{F}{m}\) = \(\frac{600}{30}\) = \(20\; m/s^2\)

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

The equation of motion can be written as:
\(F – T \) = \(m_1a\) 
\(T\) = \(F – m_1a\) 
= \(600 – 10 \times 20 \)
= \(400 \;N\) …………. (i)

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

The equation of motion can be written as:
\(F – T\) = \(m_2a\) 
\(T\) = \(F – m_2a\) 
\(T\) = \(600 – 20 × 20\) 
= \(200 \;N\) ………….. (ii) 

Answer 2

(i) Acceleration of the whole system a
\(=(\frac{F}{M_1 +M_1} = \frac {600}{10+20})\)
= 20 ms^-2
So, the net force, acting on A
⇒ 600 - T = m₁ (a)
∴ 600 - T = 10 × 20
⇒ T = 400 N.

(ii )similar to the above case, we have
So, the net force acting on B
= 600 - T = m₂ (a)
∴ 600 - T = 20 × 20
⇒ T = 200 N.