A block of mass 20kg is placed on a rough horizontal plane and a horizontal force of 12N is applied. If coefficient of friction is 0.1 the frictional force acting on it is,

Options

A. 20 N
B. 12 N
C. 8 N
D. 28 N

Answer

To find the frictional force acting on a block of mass 20 kg placed on a rough horizontal plane when a horizontal force of 12 N is applied, we can follow these steps:

### Step 1: Identify the given values
- Mass of the block (m) = 20 kg
- Applied force (F) = 12 N
- Coefficient of friction (μ) = 0.1

### Step 2: Calculate the normal force (R)
On a horizontal plane, the normal force (R) is equal to the weight of the block, which is given by:
\[ R = mg \]
Where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \).

Calculating R:
\[ R = 20 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 196 \, \text{N} \]

### Step 3: Calculate the maximum frictional force (f_max)
The maximum frictional force can be calculated using the formula:
\[ f_{\text{max}} = \mu R \]
Substituting the values:
\[ f_{\text{max}} = 0.1 \times 196 \, \text{N} = 19.6 \, \text{N} \]

### Step 4: Compare the applied force with the maximum frictional force
The applied force is 12 N, which is less than the maximum frictional force (19.6 N). This means that the frictional force will equal the applied force since friction opposes the motion.

### Step 5: Conclusion
Since the applied force (12 N) is less than the maximum frictional force (19.6 N), the frictional force acting on the block is:
\[ f = 12 \, \text{N} \]

### Final Answer
The frictional force acting on the block is **12 N**.

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