Two blocks of masses 10 kg and 20 kg are connected by a massless string and are placed on a smooth horizontal surface as shown in the figure. If a force F = 600 N is applied to 10 kg block, then the tension in the string is

Question

Smart Achievers · Accepted Answer

(4)

Here, $m_{1} = 10 kg, m_{2} = 20 kg, F = 600 N$

Let $T$ be tension of the string and $a$ be common acceleration of the system.

$∴$ $a = \frac{F}{m_{1} + m_{2}} = \frac{600 N}{10 kg + 20 kg} = \frac{600}{30} {m s}^{- 2} = 20 {m s}^{- 2}$

When a force $F$ is applied on 10 kg block, then the tension in the string is

$T = m_{2} a = (20 kg) (20 {m s}^{- 2}) = 400 N$

Solution

Q.
Two blocks of masses 10 kg and 20 kg are connected by a massless string and are placed on a smooth horizontal surface as shown in the figure. If a force F = 600 N is applied to 10 kg block, then the tension in the string is

1 100 N

2 200 N

3 300 N

4 400 N

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