Question

A rocket of.mass 1000kg containing propellant gas of 3000kh is to be launche vertically. If the fuel is consumed at a steady rate of 60kg/s, calculate the least velocity of the exhaust gases if the rocket and the content will lift off the launching pad immediately after firing?

A 15kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 200N and breaks as the elevator accelerates. What was the elevatos's minimum acceleration

Answer 1

To solve both problems, we'll go step by step.

### Problem 1: Rocket Launch

For the rocket to lift off, we need to consider the forces involved and the principle of thrust. The rocket must exert a thrust greater than its weight to start ascending.

1. **Weight of the Rocket**:
\[
\text{Weight} = m g
\]
where \( m = 1000 \, \text{kg} + 3000 \, \text{kg} = 4000 \, \text{kg} \) (total mass of the rocket and propellant) and \( g \approx 9.81 \, \text{m/s}^2 \).

\[
\text{Weight} = 4000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 39240 \, \text{N}
\]

2. **Thrust Required**:
For the rocket to lift off, the thrust \( T \) provided by the exhaust gases must be greater than the weight:
\[
T > 39240 \, \text{N}
\]

3. **Thrust Equation**:
The thrust produced by the rocket can be expressed as:
\[
T = \dot{m} v_e
\]
where \( \dot{m} = 60 \, \text{kg/s} \) (mass flow rate of the exhaust) and \( v_e \) is the velocity of the exhaust gases.

4. **Finding the Least Exhaust Velocity (\(v_e\))**:
We set the thrust equal to the weight for the rocket to just lift off:
\[
\dot{m} v_e = 39240 \, \text{N}
\]

Substituting \( \dot{m} \):
\[
60 \, \text{kg/s} \cdot v_e = 39240 \, \text{N}
\]
\[
v_e = \frac{39240 \, \text{N}}{60 \, \text{kg/s}} = 654 \, \text{m/s}
\]

Therefore, the least velocity of the exhaust gases required for the rocket to lift off is **654 m/s**.

---

### Problem 2: Elevator Acceleration and Monkey's Cord Tension

1. **Weight of the Monkey**:
The weight \( W \) of the monkey can be calculated using:
\[
W = m g
\]
where \( m = 15 \, \text{kg} \) and \( g \approx 9.81 \, \text{m/s}^2 \).
\[
W = 15 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 147.15 \, \text{N}
\]

2. **Condition for the Cord to Break**:
The tension \( T \) in the cord must satisfy:
\[
T = W + F \quad (F = ma \text{ is the force due to elevator's acceleration})
\]
Here, \( F = 15a \), where \( a \) is the elevator's acceleration.

The cord can withstand a maximum tension of \( T_{max} = 200 \, \text{N} \):
\[
200 \, \text{N} = 147.15 \, \text{N} + 15a
\]

3. **Solving for Elevator's Minimum Acceleration \( a \)**:
Rearranging the equation:
\[
15a = 200 \, \text{N} - 147.15 \, \text{N}
\]
\[
15a = 52.85 \, \text{N}
\]
\[
a = \frac{52.85 \, \text{N}}{15 \, \text{kg}} \approx 3.5233 \, \text{m/s}^2
\]

Therefore, the elevator's minimum acceleration before the cord breaks is approximately **3.52 m/s²**.