Two bodies of masses 10 kg and 20 kg respectively kept on a smooth,
Question.
Two bodies of masses $10 \mathrm{~kg}$ and $20 \mathrm{~kg}$ respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force $F=600 \mathrm{~N}$ is applied to

(i) A,

(ii) $B$ along the direction of string. What is the tension in the string in each case?

solution:

Horizontal force, F = 600 N

Mass of body $A, m_{1}=10 \mathrm{~kg}$

Mass of body B, $m_{2}=20 \mathrm{~kg}$

Total mass of the system, $m=m_{1}+m_{2}=30 \mathrm{~kg}$

Using Newton’s second law of motion, the acceleration (a) produced in the system can be calculated as:

$F=m a$

$\therefore a=\frac{F}{m}=\frac{600}{30}=20 \mathrm{~m} / \mathrm{s}^{2}$

When force $F$ is applied on body $A$ :

Two bodies of masses 10 kg and 20 kg respectively kept on a smooth,

The equation of motion can be written as:

$F-T=m_{1} a$

$\therefore T=F-m_{1} a$

$=600-10 \times 20=400 \mathrm{~N} \ldots$ (i)

When force $F$ is applied on body B:

Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface

The equation of motion can be written as:

$F-T=m_{2} a$

$T=F-m_{2} a$

$\therefore T=600-20 \times 20=200 \mathrm{~N} \ldots$ (ii)
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