Chapter 6, Problem 29P

To determine

a) Change in potential energy of hiker.

b) Minimum work required of the hiker.

c) The difference in work.

Answer to Problem 29P

Solution:a) Change in potential energy of hiker’s is $9.05\times {10}^5\text{J}.$

b) Minimum work required of the hiker is $9.05\times {10}^5\text{J}.$

c) Yes, it is more than the actual work.

Explanation of Solution

Given:Mass of hiker, $m=66.5\text{ kg}$

The height of peak is, $h_f=2660\text{ m}$

Initial height, $h_i=1270\text{ m}$

Formula Used:

Change Potential energy is given by

  $\begin{array}{l}\Delta PE=P{E}_{final}-P{E}_{initial}\\ \Delta PE=mg{h}_f-mg{h}_i\\ \Delta PE=mg(h_f-h_i)\end{array}$

Where, m is mass of the hiker, g is the acceleration due to gravity

  $h_f$ isthe height of the peak, $h_i$ is the initial height.

Calculations: a)Substitute these values in above equation and solve for change in potential energy

  $\begin{array}{l}\Delta PE=mg(h_f-h_i)\\ \Delta PE=66.5\times 9.8(2660-1270)\\ \Delta PE=905863\text{ J}\end{array}$

Change in potential energy is $9.05\times {10}^5\text{J}.$

b)

The minimum work required by the hiker to climb the peak is equal to the change in potential energy. Minimum work required for the hiker is $9.05\times {10}^5\text{J}.$

c) Yes,the actual work done by hiker may be more than this, because the hiker almost certainly had to overcome some dissipative forces such as air friction. Also, as the hiker steps up and down, they do not get the full amount of work back from each up-down event. For example, there will be friction in their joints and muscles.

Conclusion:a) Change in potential energy of hiker’s is $9.05\times {10}^5\text{J}.$

b) Minimum work required of the hiker is same as that of change in potential energy $9.05\times {10}^5\text{J}.$

c) Yes, it is more than the actual work.