Understand the setup

Two blocks of masses:

• upper block: $m = 2\,\text{kg}$
• lower block: $M = 8\,\text{kg}$

are in contact, and the table is smooth. The only friction present is between the two blocks.

We need the maximum horizontal force $F$ applied on block $M$ such that both blocks move together without slipping.

Condition for moving together

If both blocks move together, they will have the same acceleration, say $a$.

Since the table is smooth, the acceleration of the whole system is $a = \frac{F}{M+m}$

Role of friction

The upper block of mass $m$ is accelerated only by friction.

So required friction on the upper block is $f = ma = m\cdot \frac{F}{M+m}$

For no slipping, this required friction must not exceed the maximum static friction: $f \le f_{\max} = \mu_s N$

Here, $N = mg$ so $f_{\max} = \mu_s mg$

Apply the no-slip condition

$m\cdot \frac{F}{M+m} \le \mu_s mg$

Cancel $m$ from both sides: $\frac{F}{M+m} \le \mu_s g$

Therefore, $F \le (M+m)\mu_s g$

Substitute values

Given: $M=8,\quad m=2,\quad \mu_s = 0.5,\quad g=9.8\,\text{m/s}^2$

$F_{\max} = (8+2)(0.5)(9.8)$ $F_{\max} = 10 \times 0.5 \times 9.8$ $F_{\max} = 49\,\text{N}$

Check options

• A: $9.8\,\text{N}$
• B: $39.2\,\text{N}$
• C: $49\,\text{N}$
• D: $78.4\,\text{N}$

Hence the correct option is C.