A block of mass `5 kg` is lying on a rough horizontal surface. The coefficient of static and Kinetic friction are `0.3 and 0.1` and `g=10m//s^(-2)` If a horizontal force of `50N` is applied on the block, the frictional force is

Options

A. `25N`
B. `5N`
C. `10N`
D. Zero

Explanation

To solve the problem, we need to determine the frictional force acting on a block of mass 5 kg on a rough horizontal surface when a horizontal force of 50 N is applied. The coefficients of static and kinetic friction are given as 0.3 and 0.1, respectively.

### Step-by-Step Solution:

1. **Calculate the Weight of the Block**:
The weight (W) of the block can be calculated using the formula:
\[ W = m \cdot g \]
where \( m = 5 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).
\[ W = 5 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 50 \, \text{N} \]

2. **Determine the Normal Force (N)**:
Since the block is on a horizontal surface and there are no vertical forces other than the weight and the normal force, the normal force (N) is equal to the weight of the block.
\[ N = W = 50 \, \text{N} \]

3. **Calculate the Maximum Static Friction Force**:
The maximum static friction force (\( f_s \)) can be calculated using the coefficient of static friction (\( \mu_s \)):
\[ f_s = \mu_s \cdot N \]
where \( \mu_s = 0.3 \).
\[ f_s = 0.3 \cdot 50 \, \text{N} = 15 \, \text{N} \]

4. **Compare Applied Force with Maximum Static Friction**:
The applied horizontal force is 50 N. Since the applied force (50 N) is greater than the maximum static friction force (15 N), the block will start to move.

5. **Calculate the Kinetic Friction Force**:
Once the block starts moving, we need to calculate the kinetic friction force (\( f_k \)) using the coefficient of kinetic friction (\( \mu_k \)):
\[ f_k = \mu_k \cdot N \]
where \( \mu_k = 0.1 \).
\[ f_k = 0.1 \cdot 50 \, \text{N} = 5 \, \text{N} \]

### Final Answer:
The frictional force acting on the block when it is moving is **5 N**.

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