A block of 1 kg is stopped against a wall by applying a force F perpendicular to the wall. If `mu=0.2` then minimum value of F will be

Options

A. 980 N
B. 49 N
C. 98 N
D. 490 N

Answer

To solve the problem of finding the minimum force \( F \) required to stop a block of mass \( 1 \, \text{kg} \) against a wall, given that the coefficient of friction \( \mu = 0.2 \), we can follow these steps:

### Step 1: Identify the Forces Acting on the Block
The forces acting on the block are:
- The gravitational force \( mg \) acting downwards.
- The applied force \( F \) acting perpendicular to the wall.
- The frictional force \( f \) acting upwards (which prevents the block from falling).

### Step 2: Calculate the Weight of the Block
The weight \( mg \) of the block can be calculated using:
\[ mg = 1 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9.8 \, \text{N} \]

### Step 3: Understand the Role of Friction
The frictional force \( f \) that acts upwards can be expressed in terms of the normal force \( N \) (which is equal to the applied force \( F \) in this case):
\[ f = \mu N = \mu F \]
Since the block is not falling, the frictional force must balance the weight of the block:
\[ f \geq mg \]

### Step 4: Set Up the Inequality
Substituting the expression for friction into the inequality gives:
\[ \mu F \geq mg \]
This can be rewritten as:
\[ F \geq \frac{mg}{\mu} \]

### Step 5: Substitute the Known Values
Now, substituting the known values of \( mg \) and \( \mu \):
\[ F \geq \frac{9.8 \, \text{N}}{0.2} \]

### Step 6: Calculate the Minimum Force
Calculating the right-hand side:
\[ F \geq \frac{9.8}{0.2} = 49 \, \text{N} \]

### Conclusion
Thus, the minimum value of the force \( F \) required to prevent the block from falling is:
\[ \boxed{49 \, \text{N}} \]