Two bodies having the same mass 2kg each have different surface areas `50m^2` and `100 m^2` in contact with a horizontal plane. If the coefficient of friction is 0.2, the forces of friction that come into play when they are in motion will be in the ratio

Options

A. `1 : 1`
B. `1 : 2`
C. `2 : 1`
D. `1 :4`

Answer

To solve the problem, we need to understand how the force of friction is calculated and what factors it depends on. The force of friction (F_friction) can be calculated using the formula:

\[ F_{\text{friction}} = \mu \cdot N \]

where:
- \( \mu \) is the coefficient of friction
- \( N \) is the normal force acting on the body

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of each body, \( m = 2 \, \text{kg} \)
- Coefficient of friction, \( \mu = 0.2 \)
- Surface areas: Body 1 has \( A_1 = 50 \, \text{m}^2 \) and Body 2 has \( A_2 = 100 \, \text{m}^2 \)

2. **Calculate the Normal Force (N):**
The normal force \( N \) for each body is equal to its weight, which can be calculated using the formula:
\[ N = m \cdot g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \).

\[ N = 2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \]

3. **Calculate the Force of Friction for Both Bodies:**
Using the formula for friction:
\[ F_{\text{friction}} = \mu \cdot N \]

For both bodies:
\[ F_{\text{friction}} = 0.2 \cdot 19.6 \, \text{N} = 3.92 \, \text{N} \]

4. **Determine the Ratio of Forces of Friction:**
Since both bodies have the same mass and the same coefficient of friction, the force of friction acting on both bodies will be the same.

Therefore, the ratio of the forces of friction for Body 1 and Body 2 is:
\[ \text{Ratio} = \frac{F_{\text{friction, Body 1}}}{F_{\text{friction, Body 2}}} = \frac{3.92 \, \text{N}}{3.92 \, \text{N}} = 1:1 \]

### Conclusion:
The forces of friction that come into play when they are in motion will be in the ratio of **1:1**.

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