A block of mass `10kg` is placed on a rough horizontal surface having coefficient of friction `mu=0.5` . If a horizontal force of `100N` is acting on it, then acceleration of the will be.

Options

A. `0.5m//s^(2)`
B. `5m//s^(2)`
C. `10m//s^(2)`
D. `15m//s^(2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass of the block, \( m = 10 \, \text{kg} \)
- Coefficient of friction, \( \mu = 0.5 \)
- Horizontal force applied, \( F = 100 \, \text{N} \)

### Step 2: Calculate the normal force
Since the block is on a horizontal surface, the normal force \( N \) is equal to the weight of the block.
\[ N = m \cdot g \]
Assuming \( g = 10 \, \text{m/s}^2 \):
\[ N = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \]

### Step 3: Calculate the frictional force
The frictional force \( f \) can be calculated using the formula:
\[ f = \mu \cdot N \]
Substituting the values:
\[ f = 0.5 \cdot 100 \, \text{N} = 50 \, \text{N} \]

### Step 4: Determine the net force acting on the block
The net force \( F_{\text{net}} \) acting on the block is the applied force minus the frictional force:
\[ F_{\text{net}} = F - f \]
Substituting the values:
\[ F_{\text{net}} = 100 \, \text{N} - 50 \, \text{N} = 50 \, \text{N} \]

### Step 5: Calculate the acceleration of the block
Using Newton's second law, \( F = m \cdot a \), we can find the acceleration \( a \):
\[ F_{\text{net}} = m \cdot a \implies a = \frac{F_{\text{net}}}{m} \]
Substituting the values:
\[ a = \frac{50 \, \text{N}}{10 \, \text{kg}} = 5 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the block is \( 5 \, \text{m/s}^2 \).

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