To determine how much work is done in raising the climber, harness, and cable to the top of the rock face, we need to break the problem down into several steps.

1. Work Done to Raise the Climber:

The climber's mass is given as $61 \, \text{kg}$, and the gravitational force acting on her is the product of her mass and gravitational acceleration:

$F_{\text{climber}} = m_{\text{climber}} \cdot g = 61 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 597.8 \, \text{N}$

The distance the climber needs to be raised is:

$d_{\text{climber}} = 55 \, \text{m} - 14 \, \text{m} = 41 \, \text{m}$

Thus, the work done to raise the climber is:

$W_{\text{climber}} = F_{\text{climber}} \cdot d_{\text{climber}} = 597.8 \, \text{N} \times 41 \, \text{m} = 24,009.8 \, \text{J}$

2. Work Done to Raise the Cable:

The cable has a mass per unit length of $0.9 \, \text{kg/m}$, so the force due to gravity on the cable varies along its length. The total length of the cable that needs to be raised is also $41 \, \text{m}$.

The mass of the cable as a function of the height $y$ is:

$m_{\text{cable}}(y) = 0.9 \, \text{kg/m} \times y$

The weight of the cable segment at height $y$ is:

$F_{\text{cable}}(y) = m_{\text{cable}}(y) \cdot g = 0.9 \, \text{kg/m} \times y \times 9.8 \, \text{m/s}^2 = 8.82 \, y \, \text{N/m}$

The work done to lift this infinitesimal portion of the cable is:

$dW_{\text{cable}} = F_{\text{cable}}(y) \cdot dy = 8.82 \, y \, \text{dy}$

To find the total work, we integrate this expression over the height of the cable from $0$ to $41 \, \text{m}$:

$W_{\text{cable}} = \int_0^{41} 8.82 \, y \, dy = 8.82 \times \frac{y^2}{2} \bigg|_0^{41}$

$W_{\text{cable}} = 8.82 \times \frac{41^2}{2} = 8.82 \times \frac{1681}{2} = 8.82 \times 840.5 = 7409.61 \, \text{J}$

3. Total Work:

The total work done to raise the climber and the cable is the sum of the two components:

$W_{\text{total}} = W_{\text{climber}} + W_{\text{cable}} = 24,009.8 \, \text{J} + 7409.61 \, \text{J} = 31,419.41 \, \text{J}$

Thus, the work done is approximately:

$W = \frac{31,419.41}{1} \, \text{J}$

So, the final answer is:

$W = \frac{31,419.41}{1} \, \text{J}$

Would you like more details or have any questions? Here are five related questions for further practice:

1. How does the length of the cable affect the total work done?
2. What would the work be if the climber's weight were different?
3. How would the problem change if the gravitational constant was 10 m/s²?
4. How does the varying weight of the cable impact the work calculation compared to a uniform weight?
5. What if the climber had to be lowered instead of lifted?

Tip: When calculating work, always account for forces that change over distance!