A helicopter lifts a `72kg` astronaut `15 m` vertically from the ocean by means of a cable. The acceleration of the astronaut is `g/(10)`. How much work is done on the astronaut by `(g=9.8 m//s^(2))`
(a) what is the kinetic energy of the block as it passese through x=2.0m`?
(b) What is the maximum kinetic energy of the block between `x=0 and x=2.0m`?

Explanation

To solve the problem step by step, we need to calculate the work done on the astronaut by the helicopter and then find the kinetic energy at different points.

### Step 1: Calculate the gravitational force acting on the astronaut.
The gravitational force (weight) acting on the astronaut can be calculated using the formula:
\[ F_g = m \cdot g \]
Where:
- \( m = 72 \, \text{kg} \) (mass of the astronaut)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating this gives:
\[ F_g = 72 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 705.6 \, \text{N} \]

### Step 2: Determine the net force acting on the astronaut.
The astronaut is accelerating upwards with an acceleration of \( \frac{g}{10} \):
\[ a = \frac{g}{10} = \frac{9.8 \, \text{m/s}^2}{10} = 0.98 \, \text{m/s}^2 \]

Using Newton's second law, we can find the net force:
\[ F_{net} = m \cdot a = 72 \, \text{kg} \cdot 0.98 \, \text{m/s}^2 = 70.56 \, \text{N} \]

### Step 3: Calculate the tension in the cable.
The tension \( T \) in the cable can be calculated by balancing the forces acting on the astronaut:
\[ T - F_g = F_{net} \]
Thus,
\[ T = F_g + F_{net} \]
Substituting the values:
\[ T = 705.6 \, \text{N} + 70.56 \, \text{N} = 776.16 \, \text{N} \]

### Step 4: Calculate the work done on the astronaut.
The work done \( W \) by the tension force as the astronaut is lifted a distance \( S = 15 \, \text{m} \) is given by:
\[ W = T \cdot S \]
Substituting the values:
\[ W = 776.16 \, \text{N} \cdot 15 \, \text{m} = 11642.4 \, \text{J} \]

### Step 5: Kinetic energy at \( x = 2.0 \, \text{m} \).
To find the kinetic energy \( KE \) at \( x = 2.0 \, \text{m} \), we can use the work-energy principle. The work done on the astronaut from \( x = 0 \) to \( x = 2.0 \, \text{m} \) will equal the change in kinetic energy:
\[ KE = W_{net} = T \cdot x - F_g \cdot x \]
Where \( x = 2.0 \, \text{m} \):
\[ KE = (776.16 \, \text{N} \cdot 2.0 \, \text{m}) - (705.6 \, \text{N} \cdot 2.0 \, \text{m}) \]
Calculating this gives:
\[ KE = (1552.32 \, \text{J}) - (1411.2 \, \text{J}) = 141.12 \, \text{J} \]

### Step 6: Maximum kinetic energy between \( x = 0 \) and \( x = 2.0 \, \text{m} \).
The maximum kinetic energy will occur at the end of the interval, which is at \( x = 2.0 \, \text{m} \). Thus, the maximum kinetic energy is the same as calculated above:
\[ KE_{max} = 141.12 \, \text{J} \]

### Final Answers:
(a) The kinetic energy of the astronaut as it passes through \( x = 2.0 \, \text{m} \) is \( 141.12 \, \text{J} \).

(b) The maximum kinetic energy of the astronaut between \( x = 0 \) and \( x = 2.0 \, \text{m} \) is also \( 141.12 \, \text{J} \).