A block of wood mass 5kg is placed on a plane making an angle `30^(@)` with the horizontal. If the co-efficient of friction between the surface of contact of the body and plane is 0.5. What force is required to keep the body sliding down with uniform velocity

Options

A. 1.6 N
B. 4.8 N
C. 6.4 N
D. 0.33 N

Answer

To solve the problem, we need to find the force required to keep a block of wood sliding down an inclined plane at a uniform velocity. The block has a mass of 5 kg, the incline makes an angle of 30 degrees with the horizontal, and the coefficient of friction between the block and the plane is 0.5.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- The weight of the block (W) acts vertically downward:
\[ W = mg = 5 \, \text{kg} \times 10 \, \text{m/s}^2 = 50 \, \text{N} \]
- The weight can be resolved into two components:
- Perpendicular to the incline:
\[ W_{\perp} = mg \cos(\theta) = 50 \cos(30^\circ) \]
- Parallel to the incline (down the slope):
\[ W_{\parallel} = mg \sin(\theta) = 50 \sin(30^\circ) \]

2. **Calculate the Components of the Weight**:
- For \( \theta = 30^\circ \):
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin(30^\circ) = \frac{1}{2} \]
- Thus,
\[ W_{\perp} = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \, \text{N} \approx 43.3 \, \text{N} \]
\[ W_{\parallel} = 50 \times \frac{1}{2} = 25 \, \text{N} \]

3. **Calculate the Normal Force (N)**:
- The normal force is equal to the perpendicular component of the weight:
\[ N = W_{\perp} = 25\sqrt{3} \, \text{N} \]

4. **Calculate the Frictional Force (F_f)**:
- The frictional force can be calculated using the coefficient of friction (\( \mu \)):
\[ F_f = \mu N = 0.5 \times 25\sqrt{3} \approx 21.65 \, \text{N} \]

5. **Set Up the Equation for Uniform Velocity**:
- Since the block is moving with uniform velocity, the net force acting on it must be zero. Therefore, the force applied (F) plus the frictional force must balance the component of the weight acting down the slope:
\[ W_{\parallel} = F + F_f \]
- Rearranging gives:
\[ F = W_{\parallel} - F_f \]

6. **Substitute the Values**:
- Substituting the values we calculated:
\[ F = 25 \, \text{N} - 21.65 \, \text{N} \approx 3.35 \, \text{N} \]

### Final Answer:
The force required to keep the block sliding down with uniform velocity is approximately **3.35 N**.