Friction between any two surfaces in contact is the force that opposes the relative motion between them The force of limiting friction (F) between any two surfaces in contact is directly proportional to the normal reaction (R) between them `F prop R` or `F = mu R` where `mu` is coefficient of limiting friction If `theta` is angle of friction then `mu = tan theta`
A horizontal force of `1.2kg` is applied on a `1.5kg` block which rests on a horizontal surface If the coefficient of friction is `0.3` force of friction is .

Options

A. `0.45kgf`
B. `1.2kgf`
C. `1.5kgf`
D. `0.3kgf`

Explanation

To solve the problem, we need to find the force of friction acting on a 1.5 kg block resting on a horizontal surface when a horizontal force of 1.2 kgf is applied, and the coefficient of friction is 0.3.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the block (m) = 1.5 kg
- Coefficient of friction (μ) = 0.3
- Applied horizontal force (F) = 1.2 kgf

2. **Convert the Applied Force to Newtons:**
- Since 1 kgf is equivalent to 9.8 N (or approximately 10 N), we can convert the applied force:
\[ F = 1.2 \text{ kgf} \times 10 \text{ N/kgf} = 12 \text{ N} \]

3. **Calculate the Normal Reaction Force (R):**
- The normal reaction force (R) for a block resting on a horizontal surface is equal to its weight:
\[ R = m \times g = 1.5 \text{ kg} \times 10 \text{ m/s}^2 = 15 \text{ N} \]

4. **Calculate the Maximum Limiting Friction Force (F_max):**
- The maximum limiting friction force can be calculated using the formula:
\[ F_{\text{max}} = \mu \times R \]
- Substituting the values:
\[ F_{\text{max}} = 0.3 \times 15 \text{ N} = 4.5 \text{ N} \]

5. **Determine the Actual Friction Force:**
- Since the applied force (12 N) is greater than the maximum limiting friction force (4.5 N), the block will not move, and the friction force will equal the maximum limiting friction force:
\[ F_{\text{friction}} = F_{\text{max}} = 4.5 \text{ N} \]

6. **Convert the Friction Force to KGF:**
- To convert the friction force from Newtons to kgf:
\[ F_{\text{friction}} = \frac{4.5 \text{ N}}{10 \text{ N/kgf}} = 0.45 \text{ kgf} \]

### Final Answer:
The force of friction acting on the block is **0.45 kgf**.