Chapter 6, Problem 89GP

To determine

a. The work done by the hiker against gravity.

b. The average power output in watts and horsepower.

c. The rate of energy input required.

Answer to Problem 89GP

Solution:

a. The work done by the hiker against gravity is $900,000J.$

b. The average power output in watts and horsepower are 54 watts and $7.2\times 10^{-2}hp.$

c. The input power is 360 watts and $0.48\text{ hp}\text{.}$

Explanation of Solution

Given:

The mass of the hiker is $m=65\text{kg}$

Total height covered by the hiker:

  $\begin{array}{l}h=4200-2800\text{kg}\\ h=1400\text{m}\end{array}$

Time taken by the hiker to climb to the given height is:

  $\begin{array}{l}t=4.6\text{h}\\ t=4.6\times 3600\text{s}\\ t=16,560\text{s}\end{array}$

Formula used

The formula for work done against gravity is $W=mgh$ .

Where, m is the mass, g is the gravitational acceleration and h is the height.

The formula for power is $P=\frac{W}{t}$ .

Where, W is the work done and t is the time.

Calculations

Part a

The work done by the hiker against gravity is:

  $\begin{array}{l}W=mgh\\ W=65\times 9.8\times 1400\text{J}\\ W\approx 891800\text{J}\end{array}$

Part b

The average power in watts:

  $\begin{array}{l}P=\frac{W}{t}\\ P=\frac{891800}{16,560}\text{W}\\ P=54\text{watts}\end{array}$

In horsepower:

  $\begin{array}{l}\frac{P}{746}\text{hp}\\ =\frac{54}{746}\text{hp}\\ \text{=7}\text{.2}\times {\text{10}}^{-2}\text{hp}\end{array}$

Part c

Since the efficiency is 15%, therefore the output is this much percent of input.

  $P_o=0.15\text{of}{P}_{in}$

We have $P_o=54\text{W}$

Therefore, to calculate $P_{in}$:

  $\begin{array}{c}{P}_{in}=\frac{P_o}{0.15}\\ P_{in}=\frac{54}{0.15}\text{W}\\ P_{in}=360\text{W}\\ \text{=}\frac{360}{746}\text{hp}\\ \text{= 4}\text{.8 hp}\end{array}$

Conclusion:

a. The work done by the hiker against gravity is $900,000J.$

b. The average power output in watts and horsepower are 54 watts and $7.2\times 10^{-2}hp.$

c. The input power is 360 watts and $0.48\text{ hp}\text{.}$