Calculating Power From Energy

Calculating Power from Energy

Example: Calculating the Power to Climb Stairs

What is the power output for a 60.0-kg woman who runs up a 3.00 m high flight of stairs in 3.50 s, starting from rest but having a final speed of 2.00 m/s? (See the figure below.)

A woman is standing before a set of stairs with her weight shown by a vector w pointing vertically downward, which is equal to m times g. The normal force N acting on the woman is shown by a vector pointing vertically upward, which is equal to negative w. Her velocity at this point is v sub 0 equal to zero. She runs and reaches the top of the stairs at a height h with velocity v sub f. Now she possesses potential energy as well as kinetic energy labeled as K E plus P E sub g.

When this woman runs upstairs starting from rest, she converts the chemical energy originally from food into kinetic energy and gravitational potential energy. Her power output depends on how fast she does this.

Strategy and Concept

The work going into mechanical energy is \(W\text{= KE + PE}\). At the bottom of the stairs, we take both \(\text{KE}\) and \({\text{PE}}_{g}\) as initially zero; thus, \(W={\text{KE}}_{f}+{\text{PE}}_{g}=\cfrac{1}{2}{{\mathrm{mv}}_{f}}^{2}+\text{mgh}\), where \(h\) is the vertical height of the stairs. Because all terms are given, we can calculate \(W\) and then divide it by time to get power.

Solution

Substituting the expression for \(W\) into the definition of power given in the previous equation, \(P=W/t\) yields

\(P=\cfrac{W}{t}=\cfrac{\cfrac{1}{2}{{\text{mv}}_{f}}^{2}+\text{mgh}}{t}\text{.}\)

Entering known values yields

\(\begin{array}{lll}P& =& \cfrac{0.5\left(\text{60.0 kg}\right){\left(\text{2.00 m/s}\right)}^{2}+\left(\text{60.0 kg}\right)\left({\text{9.80 m/s}}^{2}\right)\left(\text{3.00 m}\right)}{\text{3.50 s}}\\ & =& \cfrac{\text{120 J}+\text{1764 J}}{\text{3.50 s}}\\ & =& \text{538 W.}\end{array}\)

Discussion

The woman does 1764 J of work to move up the stairs compared with only 120 J to increase her kinetic energy; thus, most of her power output is required for climbing rather than accelerating.

It is impressive that this woman’s useful power output is slightly less than 1 horsepower \(\left(\text{1 hp}=\text{746 W}\right)\)! People can generate more than a horsepower with their leg muscles for short periods of time by rapidly converting available blood sugar and oxygen into work output. (A horse can put out 1 hp for hours on end.) Once oxygen is depleted, power output decreases and the person begins to breathe rapidly to obtain oxygen to metabolize more food—this is known as the aerobic stage of exercise. If the woman climbed the stairs slowly, then her power output would be much less, although the amount of work done would be the same.

Making Connections: Take-Home Investigation—Measure Your Power Rating

Determine your own power rating by measuring the time it takes you to climb a flight of stairs. We will ignore the gain in kinetic energy, as the above example showed that it was a small portion of the energy gain. Don’t expect that your output will be more than about 0.5 hp.

This lesson is part of:

Work, Energy and Energy Resources

View Full Tutorial

Track Your Learning Progress

Sign in to unlock unlimited practice exams, tutorial practice quizzes, personalized weak area practice, AI study assistance with Lexi, and detailed performance analytics.