1. Understand the setup

A boy of mass $4\,\text{kg}$ stands on a wooden block of mass $5\,\text{kg}$.

• Mass of boy $= 4\,\text{kg}$
• Mass of wood $= 5\,\text{kg}$
• Coefficient of friction between wood and floor $\mu = 0.5$
• $g = 10\,\text{m/s}^2$

We need the maximum force the boy can exert on the rope such that the wood does not move.

2. Maximum friction available between wood and floor

The wood supports both its own weight and the boy's weight, so the normal reaction from the floor is

$N = (4+5)g = 9 \times 10 = 90\,\text{N}$

Hence, the maximum static friction between wood and floor is

$f_{\max} = \mu N = 0.5 \times 90 = 45\,\text{N}$

So, the floor can exert at most $45\,\text{N}$ horizontally on the wood to keep it from sliding.

3. Relation between force on rope and force on wood

Let the boy pull the rope with force $F$.

Then the rope pulls the boy upward with tension $F$.

Since the boy remains at rest relative to the wood, for the boy in vertical equilibrium:

• Upward forces: normal reaction from wood $N_1$ and rope tension $F$
• Downward force: weight $4g = 40\,\text{N}$

So,

$N_1 + F = 40$ $N_1 = 40 - F$

Thus the boy presses down on the wood with force $40 - F$.

Now consider the wood:

• Its own weight downward: $50\,\text{N}$
• Boy pushes downward on it: $40 - F$
• Rope arrangement pulls upward on the wood by tension $F$ at the point where rope is attached to the wood/pulley support.

Therefore the net load transmitted by the boy-rope interaction to the wood is reduced appropriately, but for horizontal motion the important effect is this:

In the usual arrangement, the rope exerts a horizontal pull of magnitude $F$ on the wood/pulley support, and the wood will remain at rest only if floor friction can balance it.

Hence for no motion of the wood,

$F \le f_{\max} = 45\,\text{N}$

But we must also ensure the boy stays in contact with the wood:

$N_1 = 40 - F \ge 0$ $F \le 40\,\text{N}$

So the force cannot exceed $40\,\text{N}$ due to contact condition.

4. Take the limiting value

The two conditions are:

$F \le 45\,\text{N}$ $F \le 40\,\text{N}$

Therefore,

$F_{\max} = 40\,\text{N}$

5. Final answer

$\boxed{40}$

So the maximum force is $40\,\text{N}$.

6. Comparison with stored answer

Stored correct answer: $30$

My derived answer is $40$, so I do not agree with the stored answer.

A value of $30\,\text{N}$ would arise only under a different rope geometry or interpretation not stated in the question. Under the standard interpretation, the limiting factor is the boy losing contact with the wood, giving $F_{\max}=40\,\text{N}$, while floor friction can still support up to $45\,\text{N}$.