A helicopter of mass 1000kg rises with a vertical acceleration of `15ms^-2` The crew and the passengers weigh 300kg.
Action of the rotor of the helicoper on the surrounding air

Options

A. 32500N
B. 75200N
C. 22200N
D. 23200N

Explanation

To solve the problem of finding the action of the rotor of the helicopter on the surrounding air, we can follow these steps:

### Step 1: Identify the masses involved
- Mass of the helicopter (M) = 1000 kg
- Mass of the crew and passengers (m) = 300 kg
- Total mass (Mt) = M + m = 1000 kg + 300 kg = 1300 kg

### Step 2: Calculate the weight of the helicopter with crew and passengers
- Weight (W) = Total mass (Mt) × g
- Where g (acceleration due to gravity) is approximately 10 m/s².
- Therefore, W = 1300 kg × 10 m/s² = 13000 N (downward).

### Step 3: Identify the upward acceleration of the helicopter
- The helicopter is rising with an acceleration (A) = 15 m/s².

### Step 4: Apply Newton's second law
- According to Newton's second law, the net external force (F_net) acting on the helicopter is given by:
\[ F_{net} = Mt \cdot A \]
- Here, \( F_{net} = F - W \) (where F is the force exerted by the rotor on the air).

### Step 5: Set up the equation
- From Newton's second law, we have:
\[ F - W = Mt \cdot A \]
- Rearranging gives:
\[ F = W + Mt \cdot A \]

### Step 6: Substitute the values
- Substitute W and Mt into the equation:
\[ F = 13000 N + (1300 kg \cdot 15 m/s²) \]
- Calculate \( 1300 kg \cdot 15 m/s² = 19500 N \).

### Step 7: Calculate the total force F
- Now, substitute this back into the equation:
\[ F = 13000 N + 19500 N = 32500 N \]

### Conclusion
- The action of the rotor of the helicopter on the surrounding air is **32500 N**.