Question

A block of mass \(5\) kg is moving on an inclined plane which makes an angle of \(30^\circ\) with the horizontal. The coefficient of friction between the block and the inclined plane surface is \(\dfrac{\sqrt{3}}{2}\). The force to be applied on the block so that the block moves {down the plane without acceleration is ________ N. (\( g = 10 \, \text{m s}^{-2} \))}

• A. \(7.5\)
• B. \(15\)
• C. \(25\)
• D. \(12.5\)

Hint:

For constant velocity on an inclined plane, always equate forces along the plane—acceleration is zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Moving "without acceleration" means moving at constant velocity, where the net force is zero. Friction always opposes the direction of motion.
Step 2: Detailed Explanation:
1. Forces acting on the block:
- Weight component down the incline: \(W_{||} = mg \sin 30^\circ = 5 \cdot 10 \cdot 0.5 = 25\ \text{N}\).
- Normal force: \(N = mg \cos 30^\circ = 5 \cdot 10 \cdot \frac{\sqrt{3}}{2} = 25\sqrt{3}\ \text{N}\).
- Kinetic friction (opposing downward motion): \(f_k = \mu N = \frac{\sqrt{3}}{2} \cdot 25\sqrt{3} = \frac{75}{2} = 37.5\ \text{N}\).
2. Force balance for constant downward velocity:
Let \(F\) be the applied force. Since \(f_k>W_{||}\), the applied force must be in the downward direction to assist gravity in overcoming friction.
\[ F + W_{||} = f_k \] \[ F + 25 = 37.5 \Rightarrow F = 12.5\ \text{N} \] Step 3: Final Answer:
The force to be applied is \(12.5\ \text{N}\).