A block is sliding on a rough horizontal surface. If the contact force on the block is `sqrt2` times the frictional force, the coefficient of friction is

Options

A. `0.25`
B. `(1)/(sqrt2)`
C. `sqrt2`
D. `1`

Answer

To solve the problem, we need to find the coefficient of friction (μ) given that the contact force (normal force, N) on a sliding block is √2 times the frictional force (f).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- The block is sliding on a rough horizontal surface.
- The forces acting on the block are:
- The weight of the block (mg) acting downward.
- The normal force (N) acting upward.
- The frictional force (f) acting opposite to the direction of motion.

2. **Express the Normal Force in Terms of Frictional Force**:
- According to the problem, the normal force (N) is √2 times the frictional force (f):
\[ N = \sqrt{2} \cdot f \]

3. **Relate Frictional Force to Coefficient of Friction**:
- The frictional force (f) can be expressed in terms of the coefficient of friction (μ) and the normal force (N):
\[ f = \mu \cdot N \]

4. **Substitute the Expression for Normal Force**:
- Substitute the expression for N from Step 2 into the equation for f:
\[ f = \mu \cdot (\sqrt{2} \cdot f) \]

5. **Rearrange the Equation**:
- Rearranging gives:
\[ f = \mu \sqrt{2} \cdot f \]
- If we divide both sides by f (assuming f ≠ 0):
\[ 1 = \mu \sqrt{2} \]

6. **Solve for the Coefficient of Friction (μ)**:
- Rearranging the equation gives:
\[ \mu = \frac{1}{\sqrt{2}} \]

### Final Answer:
The coefficient of friction (μ) is:
\[ \mu = \frac{1}{\sqrt{2}} \]