Ncert Exercises & Solutions

Question 5.15:

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

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:

$F = ma$
$\Rightarrow a = \frac{F}{m} = \frac{600}{30} = 20 m / s^{2}$

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

The equation of motion can be written as:

F – T = $m_{1} a$

Or, T = F – $m_{1} a$

= 800 – 20 x 10 = 400 N

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

The equation of motion can be written as:

F – T = $m_{2} a$

T = F – $m_{2} a$

T = 600 – 20 × 20 = 200 N.

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