An 80 kg person is parachuting and is experiencing a downward acceleration of `"2.8 m s"^(-2)`. The mass of the parachute is 5 kg. If the upward force on the open parachute is `kxx10^(2)N`, then what is the value of k? `("Take "g=9.8ms^(-2))`

Answer

To solve the problem, we need to analyze the forces acting on the system (the person and the parachute) and apply Newton's second law of motion.

### Step-by-Step Solution:

1. **Identify the Masses**:
- Mass of the person, \( m_1 = 80 \, \text{kg} \)
- Mass of the parachute, \( m_2 = 5 \, \text{kg} \)
- Total mass \( m = m_1 + m_2 = 80 \, \text{kg} + 5 \, \text{kg} = 85 \, \text{kg} \)

2. **Calculate the Weight of the System**:
- The weight \( W \) of the system (person + parachute) is given by:
\[ W = mg = 85 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 833 \, \text{N} \]

3. **Set Up the Force Equation**:
- The net force acting on the system can be expressed as:
\[ F_{\text{net}} = W - F \]
- Here, \( F \) is the upward force acting on the parachute, which is given as \( F = k \times 10^2 \, \text{N} \).

4. **Apply Newton's Second Law**:
- According to Newton's second law:
\[ F_{\text{net}} = ma \]
- The downward acceleration \( a \) is given as \( 2.8 \, \text{m/s}^2 \). Thus:
\[ F_{\text{net}} = 85 \, \text{kg} \times 2.8 \, \text{m/s}^2 = 238 \, \text{N} \]

5. **Combine the Equations**:
- From the force balance, we have:
\[ W - F = ma \]
- Substituting the values we calculated:
\[ 833 \, \text{N} - k \times 10^2 = 238 \, \text{N} \]

6. **Solve for \( k \)**:
- Rearranging the equation gives:
\[ k \times 10^2 = 833 \, \text{N} - 238 \, \text{N} \]
\[ k \times 10^2 = 595 \, \text{N} \]
- Dividing both sides by \( 10^2 \):
\[ k = \frac{595}{100} = 5.95 \]

### Final Answer:
The value of \( k \) is \( 5.95 \).

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