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)