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:
Show Hint
- \( \sqrt{3} - 1 \)
- \( \frac{\sqrt{3}}{2} \)
- \( \frac{\sqrt{2}}{3} \)
- \( 1 - \frac{\sqrt{3}}{2} \)
The Correct Option is A
Approach Solution - 1
Given:
- The angle of inclination, \( \theta = 60^\circ \),
- The acceleration of the block, \( a = \frac{g}{2} \),
- The gravitational acceleration, \( g \). We need to find the coefficient of kinetic friction, \( \mu_k \).
Step 1: Analyze the forces acting on the block.
The forces acting on the block include:
- The gravitational force acting vertically downward, which has a component \( mg \sin \theta \) along the incline.
- The normal force, \( N = mg \cos \theta \).
- The frictional force opposing the motion, \( F_f = \mu_k N = \mu_k mg \cos \theta \). The net force causing the block to slide down is: \[ F_{\text{net}} = mg \sin \theta - F_f \] The net force is also equal to the mass times the acceleration: \[ F_{\text{net}} = ma \]
Step 2: Set up the equation.
Equating the two expressions for \( F_{\text{net}} \): \[ mg \sin \theta - \mu_k mg \cos \theta = ma \] Since the acceleration \( a = \frac{g}{2} \), substitute this into the equation: \[ mg \sin \theta - \mu_k mg \cos \theta = m \cdot \frac{g}{2} \] Canceling out the mass \( m \) on both sides: \[ g \sin \theta - \mu_k g \cos \theta = \frac{g}{2} \]
Step 3: Simplify the equation.
Substitute \( \theta = 60^\circ \) into the equation: \[ g \sin 60^\circ - \mu_k g \cos 60^\circ = \frac{g}{2} \] Using the known values \( \sin 60^\circ = \frac{\sqrt{3}}{2} \) and \( \cos 60^\circ = \frac{1}{2} \): \[ g \cdot \frac{\sqrt{3}}{2} - \mu_k g \cdot \frac{1}{2} = \frac{g}{2} \] Dividing through by \( g \) and simplifying: \[ \frac{\sqrt{3}}{2} - \frac{\mu_k}{2} = \frac{1}{2} \] Multiplying through by 2: \[ \sqrt{3} - \mu_k = 1 \]
Step 4: Solve for \( \mu_k \).
\[ \mu_k = \sqrt{3} - 1 \]
Final Answer \sqrt{3 - 1}
Approach Solution -2
Given the equation of motion for the object on an inclined plane: \[ a = \frac{g}{2} \] The forces acting on the object are: - The gravitational force component along the plane: \( mg \sin 60^\circ \) - The frictional force: \( \mu mg \cos 60^\circ \) Using Newton's second law: \[ mg \sin 60^\circ - \mu mg \cos 60^\circ = ma \] Substitute the known acceleration: \[ g \sin 60^\circ - \mu g \cos 60^\circ = \frac{g}{2} \] Simplifying: \[ \frac{\sqrt{3}}{2} g - \frac{\mu}{2} g = \frac{1}{2} g \] Now solving for \(\mu\): \[ \mu = \sqrt{3} - 1 \] \[ \boxed{\mu = \sqrt{3} - 1} \]
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