A block of mass `m` is moving on a rough horizontal surface. `mu` is the coefficient of kinetic friction between the block and the surface. What is the net force exerted by the surface on the block?

Options

A. `mg sqrt(1+mu^(2))`
B. `mumg`
C. `mg`
D. `mgsqrt(1-mu^(2))`

Answer

To find the net force exerted by the surface on a block of mass \( m \) moving on a rough horizontal surface with a coefficient of kinetic friction \( \mu \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The block experiences several forces:
- The gravitational force acting downwards: \( F_g = mg \)
- The normal force acting upwards from the surface: \( N \)
- The applied force \( F \) (if any) acting horizontally.
- The frictional force \( f_k \) acting opposite to the direction of motion due to kinetic friction.

2. **Determine the Normal Force:**
- Since the block is on a horizontal surface and there is no vertical acceleration, the normal force \( N \) is equal to the gravitational force:
\[ N = mg \]

3. **Calculate the Frictional Force:**
- The frictional force \( f_k \) can be calculated using the coefficient of kinetic friction:
\[ f_k = \mu N = \mu mg \]

4. **Net Force Calculation:**
- The net force exerted by the surface on the block includes both the normal force and the frictional force. Since these forces act in perpendicular directions, we can find the resultant force using the Pythagorean theorem:
\[ F_{net} = \sqrt{N^2 + f_k^2} \]
- Substituting the values of \( N \) and \( f_k \):
\[ F_{net} = \sqrt{(mg)^2 + (\mu mg)^2} \]
- Factoring out \( mg \):
\[ F_{net} = mg \sqrt{1 + \mu^2} \]

5. **Final Result:**
- Therefore, the net force exerted by the surface on the block is:
\[ F_{net} = mg \sqrt{1 + \mu^2} \]