Question

A cubic block of mass $m$ is sliding down on an inclined plane at 60° with an acceleration of $\frac{g}{2}$, the value of coefficient of kinetic friction is

• A. $\sqrt{3}-1$
• B. $\frac{\sqrt{3}-1}{2}$
• C. $\frac{\sqrt{3}-1}{\sqrt{3}}$
• D. $\frac{\sqrt{3}+1}{2}$

Answer 1

Answer: (A)

$\sqrt{3}-1$

Answer 2

1. Forces on the block:
   • Weight component down the incline: $mg \sin 60^\circ$
   • Normal force: $N = mg \cos 60^\circ$
   • Kinetic friction: $f_k = \mu_k N = \mu_k mg \cos 60^\circ$
2. Newton's second law: $mg \sin 60^\circ - f_k = ma$
3. Substitute values: $mg \frac{\sqrt{3}}{2} - \mu_k mg \frac{1}{2} = m \frac{g}{2}$
4. Simplify and solve for $\mu_k$: $g \frac{\sqrt{3}}{2} - \mu_k g \frac{1}{2} = \frac{g}{2}$ $\sqrt{3} - \mu_k = 1$ $\mu_k = \sqrt{3} - 1$