A cave rescue team lifts an injured person directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of 12.0 m: (a) the initially stationary spelunker is accelerated to a speed of 5.00 m/s, (b) he is then lifted at the constant speed of 5.00 m/s, (c) finally he is decelerated to zero speed. How much work is done on the 85.0 kg rescuee by the force lifting him during each stage ?

Answer

To solve the problem of how much work is done on the injured person by the force lifting him during each stage of the rescue operation, we will analyze each stage separately.

### Given Data:
- Mass of the rescuee, \( m = 85.0 \, \text{kg} \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)
- Distance for each stage, \( s = 12.0 \, \text{m} \)
- Final speed in stage (a) and (c), \( v = 5.0 \, \text{m/s} \)

### Stage (a): Acceleration to 5.00 m/s
1. **Calculate acceleration**:
Using the equation of motion:
\[ v^2 = u^2 + 2as \]
Here, \( u = 0 \) (initial speed), \( v = 5.0 \, \text{m/s} \), and \( s = 12.0 \, \text{m} \).
\[ 5^2 = 0 + 2a(12) \implies 25 = 24a \implies a = \frac{25}{24} \approx 1.04 \, \text{m/s}^2 \]

2. **Calculate the tension in the cable**:
Using Newton's second law:
\[ T - mg = ma \implies T = mg + ma \]
\[ T = 85 \times 9.8 + 85 \times 1.04 = 833 + 88.4 = 921.4 \, \text{N} \]

3. **Calculate work done**:
Work done \( W_1 \) is given by:
\[ W_1 = T \cdot s = 921.4 \times 12 = 11057 \, \text{J} \]

### Stage (b): Constant Speed of 5.00 m/s
1. **Calculate the tension**:
When moving at constant speed, acceleration is zero:
\[ T - mg = 0 \implies T = mg \]
\[ T = 85 \times 9.8 = 833 \, \text{N} \]

2. **Calculate work done**:
Work done \( W_2 \) is:
\[ W_2 = T \cdot s = 833 \times 12 = 9996 \, \text{J} \]

### Stage (c): Deceleration to Zero Speed
1. **Calculate deceleration**:
Using the same equation of motion:
\[ 0 = (5)^2 + 2(-a)(12) \implies 25 = 24a \implies a = 1.04 \, \text{m/s}^2 \]

2. **Calculate the tension in the cable**:
\[ T - mg = -ma \implies T = mg - ma \]
\[ T = 85 \times 9.8 - 85 \times 1.04 = 833 - 88.4 = 744.6 \, \text{N} \]

3. **Calculate work done**:
Work done \( W_3 \) is:
\[ W_3 = T \cdot s = 744.6 \times 12 = 8935.2 \, \text{J} \]

### Summary of Work Done in Each Stage:
- \( W_1 \) (Stage a): \( 11057 \, \text{J} \)
- \( W_2 \) (Stage b): \( 9996 \, \text{J} \)
- \( W_3 \) (Stage c): \( 8935.2 \, \text{J} \)