A system of two blocks of masses $$m = 2$$ kg and $$M = 8$$ kg is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is $$0.5$$. The maximum horizontal force $$F$$ that can be applied to the block of mass $$M$$ so that the blocks move together will be $$(g = 9.8$$ m s$$^{-2})$$

Given:$$m=2kg,\ M=8kg,\ μ=0.5,\ g=9.8m/s^2$$

Step 1: Maximum acceleration (no slipping)

Maximum acceleration comes from maximum static friction acting on the top block:

$$f_{\max}=\mu N=\mu mg\ $$

This friction is the only horizontal force accelerating the top block:

$$⁡f_{\max}=ma_{\max}$$

$$μmg=ma_{\max}\ ​⇒\ a_{\max}​=μg=0.5\times9.8=4.9m/s^2$$

This ensures the top block does not slip over the bottom block.

Friction provides acceleration to top block:

$$a_{\max}​=μg=0.5\times9.8=4.9m/s^2$$

Step 2: Treat system as one body
Total mass:

m+M=2+8=10 kg

Step 3: Maximum force

$$F_{\max}​=(m+M)a_{\max}​=10\times4.9=49N$$

Final Answer:

49 N