A horizontal force of 1 N is needed to keep a block of mass 0.25 kg sliding on a horizontal surface with a constant speed. The coefficient of sliding friction must be : [take `g = 10 ms^(–2)`]

Options

A. 0.33
B. `0.40`
C. `0.50`
D. ` 0.25`

Answer

To solve the problem, we need to find the coefficient of sliding friction (μ) when a horizontal force of 1 N is applied to a block of mass 0.25 kg sliding on a horizontal surface at constant speed.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- The block is subjected to a horizontal force (F) of 1 N.
- The frictional force (f) opposing the motion is given by the formula:
\[ f = \mu \cdot N \]
- Here, \(N\) is the normal force acting on the block.

2. **Calculate the Normal Force**:
- The normal force \(N\) for a block resting on a horizontal surface is equal to the weight of the block.
- The weight \(W\) of the block can be calculated using the formula:
\[ W = m \cdot g \]
- Given that the mass \(m = 0.25 \, \text{kg}\) and \(g = 10 \, \text{m/s}^2\):
\[ N = W = 0.25 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 2.5 \, \text{N} \]

3. **Set Up the Equation for Friction**:
- Since the block is moving at a constant speed, the net force acting on it is zero. Therefore, the applied force is equal to the frictional force:
\[ F = f \]
- Substituting the known values:
\[ 1 \, \text{N} = \mu \cdot N \]

4. **Substitute the Normal Force into the Equation**:
- We already calculated \(N = 2.5 \, \text{N}\):
\[ 1 \, \text{N} = \mu \cdot 2.5 \, \text{N} \]

5. **Solve for the Coefficient of Friction (μ)**:
- Rearranging the equation gives:
\[ \mu = \frac{1 \, \text{N}}{2.5 \, \text{N}} = \frac{1}{2.5} = 0.4 \]

### Final Answer:
The coefficient of sliding friction (μ) is **0.4**.

---