Question

A block of certain mass is placed on a rough inclined plane. The angle between the plane and the horizontal is $30^0$. The coefficients of static and kinetic frictions between the block and the inclined plane are 0.6 and 0.5 respectively. Then the magnitude of the acceleration of the block is [Take g=10 ms-2]

• A. $2 ms^{-2}$
• B. Zero
• C. $0.196 ms^{-2}$
• D. $0.67 ms^{-2}$

Answer 1

The Correct Option is B

Given:
The angle of the inclined plane \( \theta = 30^\circ \)
Coefficients of static friction \( \mu_s = 0.6 \) and kinetic friction \( \mu_k = 0.5 \)
Gravitational acceleration \( g = 10 \, \text{m/s}^2 \) ### Forces acting on the block: 1.
The component of gravitational force along the incline: \[ F_{\text{gravity, parallel}} = mg \sin \theta \] 2. The component of the normal force: \[ F_{\text{normal}} = mg \cos \theta \] 3. The static frictional force opposing the motion: \[ F_{\text{friction, static}} = \mu_s F_{\text{normal}} = \mu_s mg \cos \theta \]Condition for the block to remain at rest: For the block to remain at rest, the maximum static friction force must balance the component of the gravitational force along the incline. The maximum static frictional force is: \[ F_{\text{friction, static max}} = \mu_s mg \cos \theta \] Substitute the values: \[ F_{\text{friction, static max}} = 0.6 \times m \times 10 \times \cos 30^\circ \] \[ F_{\text{friction, static max}} = 0.6 \times m \times 10 \times \frac{\sqrt{3}}{2} \] \[ F_{\text{friction, static max}} = 3 \sqrt{3} \, m \] Now, the gravitational force along the incline is: \[ F_{\text{gravity, parallel}} = m \times 10 \times \sin 30^\circ = 5 m \] Since the static frictional force \( F_{\text{friction, static max}} = 3 \sqrt{3} m \) is greater than the component of gravity along the incline, the block will not move. The net force is zero.
Since the frictional force is enough to balance the gravitational pull, the block does not move and the acceleration is zero.

Thus, the correct answer is (B) zero.

Answer 2

We begin by analyzing the forces acting on the block on the inclined plane. The forces involved are:
1. Gravitational force (\(mg\)) acting vertically downward.
2. Normal force (\(N\)) acting perpendicular to the surface of the inclined plane.
3. Frictional force (\(f\)), which opposes the motion of the block.

The force components are as follows:
The component of gravitational force along the plane: \[ F_{\text{gravity parallel}} = mg \sin \theta \]
The component of the gravitational force perpendicular to the plane: \[ F_{\text{gravity perpendicular}} = mg \cos \theta \]
The frictional force is given by: \[ f_{\text{friction}} = \mu N \] where \(\mu\) is the coefficient of friction, and \(N\) is the normal force. For kinetic friction, \(\mu = 0.5\), and the normal force is \(N = mg \cos \theta\). Substituting the values for \(g = 10 \, \text{m/s}^2\) and \(\theta = 30^\circ\):
1. The component of gravitational force parallel to the incline is: \[ F_{\text{gravity parallel}} = m \times 10 \times \sin 30^\circ = \frac{m \times 10}{2} = 5m \] 2. The normal force is: \[ N = m \times 10 \times \cos 30^\circ = m \times 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}m \] 3. The frictional force is: \[ f_{\text{friction}} = 0.5 \times 5\sqrt{3}m = 2.5\sqrt{3}m \approx 4.33m \]
Now, compare the frictional force with the gravitational component parallel to the plane: 
The gravitational force parallel to the incline is \(5m\).
The frictional force is \(4.33m\). Since the frictional force exceeds the gravitational force component, the block does not move, and the acceleration of the block is zero.

Thus, the correct answer is (B) zero.