A horizontal force of 150N produces an acceleration of 2m/s2 in a body placed on a 6 horizontal surface. A horizontal force of 200N produces an acceleration of `3m//s^2`. The mass of the body and coefficient of kinetic friction are (g = 10 `ms^(-2)`)

Options

A. 50 kg , 0.1
B. 25 kg , 0.1
C. 50 kg , 0.5
D. 50 kg , 0.2

Answer

To solve the problem, we need to find the mass of the body and the coefficient of kinetic friction (μ). We will use Newton's second law and the equations of motion with friction.

### Step-by-Step Solution:

1. **Identify the forces acting on the body:**
- When a force \( F \) is applied to the body, it overcomes the frictional force \( f \) and produces an acceleration \( a \).
- The forces acting on the body can be represented as:
\[ F - f = ma \]
- The frictional force can be expressed as:
\[ f = \mu mg \]
- Here, \( m \) is the mass of the body, \( g \) is the acceleration due to gravity (10 m/s²), and \( \mu \) is the coefficient of kinetic friction.

2. **Set up equations for the two scenarios:**
- For the first scenario (150 N force, 2 m/s² acceleration):
\[ 150 - \mu mg = m \cdot 2 \quad \text{(1)} \]
- For the second scenario (200 N force, 3 m/s² acceleration):
\[ 200 - \mu mg = m \cdot 3 \quad \text{(2)} \]

3. **Substitute \( f \) in terms of \( m \) and \( \mu \):**
- From equation (1):
\[ 150 - \mu mg = 2m \]
- Rearranging gives:
\[ \mu mg = 150 - 2m \quad \text{(3)} \]
- From equation (2):
\[ 200 - \mu mg = 3m \]
- Rearranging gives:
\[ \mu mg = 200 - 3m \quad \text{(4)} \]

4. **Equate equations (3) and (4):**
- Setting the right-hand sides equal to each other:
\[ 150 - 2m = 200 - 3m \]
- Rearranging gives:
\[ 3m - 2m = 200 - 150 \]
\[ m = 50 \, \text{kg} \]

5. **Substitute \( m \) back to find \( \mu \):**
- Substitute \( m = 50 \) into equation (3):
\[ \mu \cdot 50 \cdot 10 = 150 - 2 \cdot 50 \]
\[ 500\mu = 150 - 100 \]
\[ 500\mu = 50 \]
\[ \mu = \frac{50}{500} = 0.1 \]

### Final Results:
- The mass of the body \( m = 50 \, \text{kg} \)
- The coefficient of kinetic friction \( \mu = 0.1 \)