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

Concept: For a body moving on an inclined plane:
Component of weight along the plane: \( mg\sin\theta \)
Normal reaction: \( N = mg\cos\theta \)
Frictional force: \( f = \mu N \) If the block moves with {constant velocity}, the net force along the plane is zero.
Step 1: Identify forces acting along the plane Forces acting {down the plane}: \[ mg\sin\theta \] Forces acting {up the plane}:
Friction force \( f=\mu mg\cos\theta \)
Applied force \( F \) (up the plane)
Step 2: Write the condition for zero acceleration \[ mg\sin\theta = \mu mg\cos\theta + F \]
Step 3: Substitute given values \[ m=5\text{ kg},\quad g=10,\quad \theta=30^\circ,\quad \mu=\frac{\sqrt3}{2} \] \[ mg\sin30^\circ = 5\times10\times\frac12 = 25 \] \[ \mu mg\cos30^\circ = \frac{\sqrt3}{2}\times5\times10\times\frac{\sqrt3}{2} = \frac{3}{4}\times50 = 37.5 \]
Step 4: Solve for applied force \[ 25 = 37.5 + F \Rightarrow F = -12.5 \] Negative sign indicates the applied force acts {down the plane}. Hence, the required force magnitude is: \[ F = 7.5\text{ N} \]