An engine of mass` 5 xx 10^4` kg pulls a coach of mass `4xx10^4kg.` Suppose that there is a resistance of 1 N per 100 kg acting on both coach and engine, and that the driving force is 4500N. The acceleration of the engine and the coupling will respectively be

Options

A. (a)`0.04 m//s^(2), 2000N`
B. (b)`0.4 m//s^(2), 200N`
C. (c)`0.4 m//s^(2), 20N`
D. (d)`4 m//s^(2) , 200N`

Answer

To solve the problem step by step, we will analyze the forces acting on both the engine and the coach, and apply Newton's second law of motion.

### Step 1: Identify the Given Data
- Mass of the engine, \( m_e = 5 \times 10^4 \, \text{kg} \)
- Mass of the coach, \( m_c = 4 \times 10^4 \, \text{kg} \)
- Total mass, \( m_{total} = m_e + m_c = 9 \times 10^4 \, \text{kg} \)
- Driving force, \( F_{drive} = 4500 \, \text{N} \)
- Resistance per 100 kg = 1 N
- Total resistance \( R \) acting on both engine and coach.

### Step 2: Calculate Total Resistance
The total resistance acting on the system can be calculated as follows:
\[ R = \frac{m_{total}}{100} = \frac{9 \times 10^4}{100} = 900 \, \text{N} \]

### Step 3: Calculate Individual Resistances
- Resistance on the engine:
\[ R_e = \frac{m_e}{100} = \frac{5 \times 10^4}{100} = 500 \, \text{N} \]
- Resistance on the coach:
\[ R_c = \frac{m_c}{100} = \frac{4 \times 10^4}{100} = 400 \, \text{N} \]

### Step 4: Set Up Equations Using Newton's Second Law
For the coach, the net force can be expressed as:
\[ T - R_c = m_c \cdot a \quad \text{(1)} \]
For the engine, the net force can be expressed as:
\[ F_{drive} - R_e - T = m_e \cdot a \quad \text{(2)} \]

### Step 5: Substitute Known Values into the Equations
Substituting the known values into the equations:
1. From equation (1):
\[ T - 400 = 4 \times 10^4 \cdot a \quad \text{(1)} \]
2. From equation (2):
\[ 4500 - 500 - T = 5 \times 10^4 \cdot a \quad \text{(2)} \]
This simplifies to:
\[ 4000 - T = 5 \times 10^4 \cdot a \quad \text{(2)} \]

### Step 6: Solve the Equations
Now we can solve equations (1) and (2) simultaneously. First, let's express \( T \) from equation (1):
\[ T = 400 + 4 \times 10^4 \cdot a \quad \text{(3)} \]
Now substitute equation (3) into equation (2):
\[ 4000 - (400 + 4 \times 10^4 \cdot a) = 5 \times 10^4 \cdot a \]
Simplifying this gives:
\[ 4000 - 400 - 4 \times 10^4 \cdot a = 5 \times 10^4 \cdot a \]
\[ 3600 = 9 \times 10^4 \cdot a \]
Now solve for \( a \):
\[ a = \frac{3600}{9 \times 10^4} = \frac{3600}{90000} = 0.04 \, \text{m/s}^2 \]

### Step 7: Find the Tension (Coupling Force)
Now substitute the value of \( a \) back into equation (3) to find \( T \):
\[ T = 400 + 4 \times 10^4 \cdot 0.04 \]
\[ T = 400 + 1600 = 2000 \, \text{N} \]

### Final Answer
The acceleration of the engine and the coupling force (tension) are:
- Acceleration \( a = 0.04 \, \text{m/s}^2 \)
- Tension \( T = 2000 \, \text{N} \)