A climber weighs 645N. Calculate the work done on the backpack, the work done lifting herself and the backpack, and her power output, assuming a backpack weight and climb height.

To solve these questions, I need the following information, which is not provided in the image:

• The weight of the backpack.
• The vertical distance (height) the climber ascends.
• The time taken for the climb.

Please provide these values so I can calculate the answers.

However, I can outline the formulas and steps you would use once you have the missing information:

Let:

• $F_{\text{backpack}}$ be the weight of the backpack (in Newtons, N).
• $h$ be the vertical height climbed (in meters, m).
• $t$ be the time taken for the climb (in seconds, s).
• $F_{\text{climber}} = 645 \, \text{N}$ (given).

(i) How much work does the climber do on the backpack?
Work done ($W$) is calculated as the force applied multiplied by the distance moved in the direction of the force. In this case, the force is the weight of the backpack, and the distance is the height climbed.

$$W_{\text{backpack}} = F_{\text{backpack}} \times h$$

• Missing: $F_{\text{backpack}}$ and $h$.

(ii) If the climber weighs 645N, how much work does she do to lift herself and the backpack?
The total force to be lifted is the sum of the climber's weight and the backpack's weight.

$$W_{\text{total}} = (F_{\text{climber}} + F_{\text{backpack}}) \times h$$ $$W_{\text{total}} = (645 \, \text{N} + F_{\text{backpack}}) \times h$$

• Missing: $F_{\text{backpack}}$ and $h$.

(iii) What is the average power developed by the climber?
Average power ($P$) is the total work done divided by the time taken.

$$P_{\text{average}} = \frac{W_{\text{total}}}{t}$$ $$P_{\text{average}} = \frac{(645 \, \text{N} + F_{\text{backpack}}) \times h}{t}$$

• Missing: $F_{\text{backpack}}$, $h$, and $t$.