Physics - NLM Question with Solution | TestHub

For $\theta \le {\theta }_R$ there is no slipping, so the contact force balances the weight: $F_c=mg=10$ N, giving $m=1$ kg - (A)

Slipping starts at ${\theta }_R=45^{\circ }$ (graph kink): $\tan {\theta }_R={\mu }_s=1$ - (C) correct, (B) wrong.

For $\theta >{\theta }_R$: $F_c=mg\cos \theta \sqrt{1+{\mu }^2}$; at $\theta =60^{\circ }$: $x=10\times \frac{1}{2}\times \sqrt{2}=5\sqrt{2}$ N, not $5\sqrt{5}$ - (D) wrong.