For m₁ mass :

m₁g - T = m₁a .........(1)

For m₂ mass :

T - m₂g = m₂a ....... (2)

By solving (1) and (2) we get,

Acceleration a = $$\left( {{{{m_1} - {m_2}} \over {{m_1} + {m_2}}}} \right)g$$

$$ = {{\left( {5 - 4.8} \right) \times 9.8} \over {\left( {5 + 4.8} \right)}}m/{s^2}$$ = 0.2 m/s²

Assume weight of the block = $$W$$ and friction force = $$f$$ = $$\mu N$$

For the block to remain stationary with the wall, weight of the block should balanced by the force of friction.

$$\therefore$$ $$f=W$$

$$ \Rightarrow $$ $$\mu N = W$$

$$ \Rightarrow $$ $$W$$ = 0.2 $$ \times $$ 10 = 2 N

As shown in the figure, the three forces are represented by the sides of a triangle taken in the same order. So the particle will be in equilibrium under the three forces.

Therefore the resultant force is zero.

We know, $${\overrightarrow F_{net}} = m\overrightarrow a .$$

$$\therefore$$ 0 = $$ m\overrightarrow a .$$

$$ \Rightarrow \overrightarrow a = 0$$

Therefore acceleration will be zero.

Hence the particle velocity remains unchanged.

Here, thrust force is responsible to accelerate the rocket,

So initial thrust of the blast

= (Lift-off mass) × acceleration

= (3.5 × 10⁴) × (10)

= 3.5 × 10⁵ N