Question

A brick of mass 2 kg slides down an incline of height 5 m and angle \( 30^\circ \). If \( \mu = \frac{1}{2\sqrt{3}} \), the velocity at the bottom is (Take \( g = 10 \text{ m/s}^2 \)) (A) 5 m/s

• A. 50 m/s
• B. 7 m/s
• C. 0
• D. 10 m/s
• E.

Hint:

Use energy method for faster solving.

Answer 1

The Correct Option is C

Concept: Work–Energy principle with friction
The loss in potential energy is partly used to do work against friction and the rest appears as kinetic energy: \[ mgh - f d = \frac{1}{2}mv^2 \]

Step 1: Calculate distance along incline
Given $\theta = 30^\circ$: \[ d = \frac{h}{\sin 30^\circ} = \frac{5}{1/2} = 10 \, \text{m} \]

Step 2: Calculate frictional force
Normal reaction: \[ N = mg\cos 30^\circ \] Friction: \[ f = \mu N = \mu mg\cos 30^\circ \] Given values: \[ f = 5 \, \text{N} \]

Step 3: Apply energy equation
Potential energy: \[ mgh = 10 \times 10 = 100 \, \text{J} \] Work done against friction: \[ f d = 5 \times 10 = 50 \, \text{J} \]

Step 4: Find kinetic energy
\[ \frac{1}{2}mv^2 = 100 - 50 = 50 \] \[ v^2 = 50 \]

Step 5: Final velocity
\[ v = \sqrt{50} \approx 7 \, \text{m/s} \] Final Answer:
\[ \boxed{v \approx 7 \, \text{m/s}} \] Conclusion:
Friction reduces the final speed by dissipating part of the mechanical energy as heat.