A helicopter of mass 1000kg rises with a vertical acceleration of `15ms^-2` The crew and the passengers weigh 300kg.
Force on the floor by the crew and passengers

Options

A. 7500N
B. 1500N
C. 2500N
D. 5000N

Explanation

To solve the problem of the force exerted on the floor by the crew and passengers in a helicopter that is rising with a vertical acceleration, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the crew and passengers (m) = 300 kg
- Acceleration due to gravity (g) = 10 m/s² (assuming standard value)
- Vertical acceleration of the helicopter (a) = 15 m/s²

2. **Calculate the Weight of the Crew and Passengers:**
- The weight (W) can be calculated using the formula:
\[ W = m \cdot g \]
- Substituting the values:
\[ W = 300 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 3000 \, \text{N} \]

3. **Determine the Net Force Acting on the Crew and Passengers:**
- According to Newton's second law, the net force (F_net) acting on the system is given by:
\[ F_{\text{net}} = m \cdot a \]
- Substituting the values:
\[ F_{\text{net}} = 300 \, \text{kg} \cdot 15 \, \text{m/s}^2 = 4500 \, \text{N} \]

4. **Calculate the Normal Force (N):**
- The normal force (N) exerted by the floor on the crew and passengers can be expressed as:
\[ N - W = F_{\text{net}} \]
- Rearranging gives:
\[ N = W + F_{\text{net}} \]
- Substitute the values of W and F_net:
\[ N = 3000 \, \text{N} + 4500 \, \text{N} = 7500 \, \text{N} \]

5. **Conclusion:**
- The force on the floor by the crew and passengers is **7500 N**.

### Final Answer:
The force on the floor by the crew and passengers is **7500 N**.

---