A fire man has to carry an injured person of mass 40 kg from the top of a building with the help of the rope which can withstand a load of 100 kg. The acceleration of the fireman if his mass is 80 kg, will be-

Options

A. `8.17 m//s^(2)`
B. `9.8 m//s^(2)`
C. `1.63 m//s^(2)`
D. `17.97 m//s^(2)`

Answer

To solve the problem, we need to analyze the forces acting on the fireman and the injured person while they are being lowered down by the rope.

### Step-by-Step Solution:

1. **Identify the Masses:**
- Mass of the injured person (m1) = 40 kg
- Mass of the fireman (m2) = 80 kg
- Total mass (M) = m1 + m2 = 40 kg + 80 kg = 120 kg

2. **Convert Mass to Weight:**
- The weight of the injured person (W1) = m1 * g = 40 kg * 9.8 m/s² = 392 N
- The weight of the fireman (W2) = m2 * g = 80 kg * 9.8 m/s² = 784 N
- Total weight (W_total) = W1 + W2 = 392 N + 784 N = 1176 N

3. **Determine the Maximum Tension in the Rope:**
- The maximum load the rope can withstand = 100 kg
- Maximum tension (T_max) = 100 kg * g = 100 kg * 9.8 m/s² = 980 N

4. **Apply Newton's Second Law:**
- The net force (F_net) acting on the system when moving downward is given by:
\[ F_{net} = W_{total} - T_{max} \]
- Substitute the values:
\[ F_{net} = 1176 N - 980 N = 196 N \]

5. **Calculate the Acceleration of the System:**
- According to Newton's second law, F_net = M * a, where M is the total mass and a is the acceleration.
- Rearranging gives:
\[ a = \frac{F_{net}}{M} \]
- Substitute the values:
\[ a = \frac{196 N}{120 kg} \approx 1.633 \, \text{m/s}^2 \]

6. **Conclusion:**
- The acceleration of the fireman while carrying the injured person is approximately **1.633 m/s²** downward.