Physics - NLM Question with Solution | TestHub

For the hanging block of mass $m$ to be in equilibrium, the tension in the string must balance its weight.$$T=mg=2\text{ kg}\times 10{\text{ m/s}}^2=20\text{ N}$$

For the block$M$ to remain at rest, the net horizontal force on it must be zero. The forces acting horizontally on $M$ are the applied force $F$, the tension $T$, and the static friction $f_s$. When $F$ is maximum, the block $M$ will be on the verge of sliding to the left, meaning friction acts to the right.$$F-T-f_s=0⟹F=T+f_s$$ The maximum static friction is$f_{s,max}={\mu }_{s}N$. Since the surface is horizontal, the normal force $N=Mg=10\text{ kg}\times 10{\text{ m/s}}^2=100\text{ N}$.$$f_{s,max}=0.4\times 100\text{ N}=40\text{ N}$$ Therefore, the maximum force$F$ is:$$F_{max}=T+f_{s,max}=20\text{ N}+40\text{ N}=60\text{ N}$$