A block of mass 0.1 is held against a wall applying a horizontal force of 5N on block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is:

Options

A. 2.5N
B. 0.98N
C. 4.9N
D. 0.49N

Explanation

To find the magnitude of the frictional force acting on the block, we can follow these steps:

### Step 1: Identify the forces acting on the block
- The block has a mass \( m = 0.1 \, \text{kg} \).
- A horizontal force \( F = 5 \, \text{N} \) is applied to the block against the wall.
- The weight of the block acting downwards is given by \( W = mg \), where \( g \approx 9.8 \, \text{m/s}^2 \).

### Step 2: Calculate the weight of the block
\[ W = mg = 0.1 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 0.98 \, \text{N} \]

### Step 3: Determine the normal force
- The normal force \( N \) exerted by the wall on the block is equal to the applied horizontal force \( F \).
\[ N = F = 5 \, \text{N} \]

### Step 4: Calculate the maximum static frictional force
- The maximum static frictional force \( f_{\text{max}} \) can be calculated using the coefficient of friction \( \mu \) and the normal force \( N \).
\[ f_{\text{max}} = \mu N = 0.5 \times 5 \, \text{N} = 2.5 \, \text{N} \]

### Step 5: Determine the actual frictional force
- The actual frictional force \( f \) acting on the block will be equal to the weight of the block since the block is not moving vertically. Thus, the frictional force must counteract the weight.
\[ f = W = 0.98 \, \text{N} \]

### Step 6: Conclusion
- The magnitude of the frictional force acting on the block is \( 0.98 \, \text{N} \).

### Final Answer
\[ \text{The magnitude of the frictional force acting on the block is } 0.98 \, \text{N}. \]

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