JEE Main 2022 — Laws Of Motion Question with Solution

Here we will use the Newton's laws of motion.

Let acceleration of 100 kg block w.r.t. ground is a.

Now, Let's draw FBD of 100 kg block w.r.t. ground,

By NLM,

$F-T-N_1=100a\text{... (1)}$

FBD of 20 kg block w.r.t. 100 kg block,

By NLM,

$$N_1=m_{2}a=20a\Rightarrow N_1=20a$$ ... (2)

$$T-m_{2}g=m_2(2)$$

$$\Rightarrow T-20g=20(2)$$

$$\Rightarrow T=40+200=240$$

$$\Rightarrow T=240N$$ .... (3)

Now, FBD of 10 kg block w.r.t. 100 kg block,

By NLM,

$$m_{1}a-T=m_1(2)$$

$$\Rightarrow 10a-T=20$$

$$\Rightarrow 10a-240=20$$ [from (3)]

$$\Rightarrow 10a=260$$

$$\Rightarrow a=26m{s}^{-2}$$

So, from (2), $$N_1=20(26)=520N$$

Now, from (1), $$F-240-520=100(26)$$

$$\Rightarrow F=2600+240+520$$

$$\Rightarrow F=3360N$$