A force of `150N` produces an acceleration of `2ms^(-2)` in a body and a force of `200N` produces an acceleration of `3ms^(-2)` The mass of the body and the coefficinent of kinetic friction are .

Options

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

Explanation

To solve the problem, we need to find the mass of the body and the coefficient of kinetic friction based on the forces and accelerations provided. Let's break it down step by step.

### Step 1: Set Up the Equations

We have two scenarios:

1. **Case 1:** A force of \( F_1 = 150 \, \text{N} \) produces an acceleration \( a_1 = 2 \, \text{m/s}^2 \).
2. **Case 2:** A force of \( F_2 = 200 \, \text{N} \) produces an acceleration \( a_2 = 3 \, \text{m/s}^2 \).

In both cases, we need to account for the force of friction, which can be expressed as \( F_{\text{friction}} = \mu m g \), where \( \mu \) is the coefficient of kinetic friction, \( m \) is the mass of the body, and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 2: Write the Equations of Motion

From Newton's second law, we can write the equations for both cases:

**For Case 1:**
\[ F_1 - F_{\text{friction}} = m a_1 \]
Substituting for \( F_{\text{friction}} \):
\[ 150 - \mu m g = m \cdot 2 \]
This simplifies to:
\[ 150 - \mu m \cdot 10 = 2m \quad \text{(Equation 1)} \]

**For Case 2:**
\[ F_2 - F_{\text{friction}} = m a_2 \]
Substituting for \( F_{\text{friction}} \):
\[ 200 - \mu m g = m \cdot 3 \]
This simplifies to:
\[ 200 - \mu m \cdot 10 = 3m \quad \text{(Equation 2)} \]

### Step 3: Rearranging the Equations

Now, we can rearrange both equations to isolate the friction term.

**From Equation 1:**
\[ 150 - 2m = \mu m \cdot 10 \]
\[ \mu m \cdot 10 = 150 - 2m \quad \text{(1)} \]

**From Equation 2:**
\[ 200 - 3m = \mu m \cdot 10 \]
\[ \mu m \cdot 10 = 200 - 3m \quad \text{(2)} \]

### Step 4: Set the Two Expressions for Friction Equal

Since both expressions equal \( \mu m \cdot 10 \), we can set them equal to each other:
\[ 150 - 2m = 200 - 3m \]

### Step 5: Solve for Mass \( m \)

Rearranging the equation:
\[ 3m - 2m = 200 - 150 \]
\[ m = 50 \, \text{kg} \]

### Step 6: Substitute \( m \) Back to Find \( \mu \)

Now that we have \( m \), we can substitute it back into either Equation 1 or Equation 2 to find \( \mu \). Let's use Equation 1:
\[ \mu \cdot 50 \cdot 10 = 150 - 2 \cdot 50 \]
\[ \mu \cdot 500 = 150 - 100 \]
\[ \mu \cdot 500 = 50 \]
\[ \mu = \frac{50}{500} = 0.1 \]

### Final Answers

- Mass of the body, \( m = 50 \, \text{kg} \)
- Coefficient of kinetic friction, \( \mu = 0.1 \)