Question

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

• A. \( \sqrt{3} - 1 \)
• B. \( \frac{\sqrt{3}}{2} \)
• C. \( \frac{\sqrt{2}}{3} \)
• D. \( 1 - \frac{\sqrt{3}}{2} \)

Hint:

The coefficient of kinetic friction can be found by analyzing the forces acting on the object and using the equation for the net force along the incline.

Answer 1

The Correct Option is A

Given:
- Angle of inclination, \( \theta = 60^\circ \) 
- Block acceleration, \( a = \frac{g}{2} \) 
- Gravitational acceleration, \( g \). Determine the coefficient of kinetic friction, \( \mu_k \). 
Step 1: Force Analysis.
Forces acting on the block are: 
- Gravitational force component along the incline: \( mg \sin \theta \). 
- Normal force: \( N = mg \cos \theta \). 
- Frictional force opposing motion: \( F_f = \mu_k N = \mu_k mg \cos \theta \). The net force driving the block down is: \[ F_{\text{net}} = mg \sin \theta - F_f \] This net force also equals mass times acceleration: \[ F_{\text{net}} = ma \]

Step 2: Equation Formulation.
Equating the expressions for \( F_{\text{net}} \): \[ mg \sin \theta - \mu_k mg \cos \theta = ma \] Substituting \( a = \frac{g}{2} \): \[ mg \sin \theta - \mu_k mg \cos \theta = m \cdot \frac{g}{2} \] Canceling mass \( m \) from both sides:
\[ g \sin \theta - \mu_k g \cos \theta = \frac{g}{2} \]

Step 3: Equation Simplification.
Substituting \( \theta = 60^\circ \):
\[ g \sin 60^\circ - \mu_k g \cos 60^\circ = \frac{g}{2} \]

Using \( \sin 60^\circ = \frac{\sqrt{3}}{2} \) and \( \cos 60^\circ = \frac{1}{2} \):
\[ g \cdot \frac{\sqrt{3}}{2} - \mu_k g \cdot \frac{1}{2} = \frac{g}{2} \]

Dividing by \( g \) and simplifying:
\[ \frac{\sqrt{3}}{2} - \frac{\mu_k}{2} = \frac{1}{2} \]

Multiplying by 2:
\[ \sqrt{3} - \mu_k = 1 \]

Step 4: Solving for \( \mu_k \).
\[ \mu_k = \sqrt{3} - 1 \]

Final Answer: \( \sqrt{3} - 1 \)