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  • Calculate power by calculating changes in energy over time.
  • Examine power consumption and calculations of the cost of energy consumed.

What is power?

Power —the word conjures up many images: a professional football player muscling aside his opponent, a dragster roaring away from the starting line, a volcano blowing its lava into the atmosphere, or a rocket blasting off, as in [link] .

A space shuttle rocket is being launched and is burning propellant.
This powerful rocket on the Space Shuttle Endeavor did work and consumed energy at a very high rate. (credit: NASA)

These images of power have in common the rapid performance of work, consistent with the scientific definition of power    ( P size 12{P} {} ) as the rate at which work is done.

Power

Power is the rate at which work is done.

P = W t size 12{P= { {W} over {t} } } {}

The SI unit for power is the watt    ( W size 12{W} {} ), where 1 watt equals 1 joule/second ( 1 W = 1 J/s ) . size 12{ \( 1" W"=1" J/s" \) "." } {}

Because work is energy transfer, power is also the rate at which energy is expended. A 60-W light bulb, for example, expends 60 J of energy per second. Great power means a large amount of work or energy developed in a short time. For example, when a powerful car accelerates rapidly, it does a large amount of work and consumes a large amount of fuel in a short time.

Calculating power from energy

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 [link] .)

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 = KE + PE size 12{W"= KE + PE"} {} . At the bottom of the stairs, we take both KE size 12{"KE"} {} and PE g as initially zero; thus, W = KE f + PE g = 1 2 mv f 2 + mgh size 12{W="KE" rSub { size 8{f} } +"PE" rSub { size 8{g} } = { { size 8{1} } over { size 8{2} } } ital "mv" rSub { size 8{f} rSup { size 8{2} } } + ital "mgh"} {} , where h size 12{h} {} is the vertical height of the stairs. Because all terms are given, we can calculate W size 12{W} {} and then divide it by time to get power.

Solution

Substituting the expression for W size 12{W} {} into the definition of power given in the previous equation, P = W / t size 12{P= {W} slash {t} } {} yields

P = W t = 1 2 mv f 2 + mgh t . size 12{P= { {W} over {t} } = { { { {1} over {2} } ital "mv" rSub { size 8{f} rSup { size 8{2} } } + ital "mgh"} over {t} } "." } {}

Entering known values yields

P = 0.5 60.0 kg 2.00 m/s 2 + 60.0 kg 9.80 m/s 2 3.00 m 3.50 s = 120 J + 1764 J 3.50 s = 538 W. alignl { stack { size 12{P= { {0 "." 5 left ("60" "." 0" kg" right ) left (2 "." "00"" m/s" right ) rSup { size 8{2} } + left ("60" "." 0" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right ) left (3 "." "00"" m" right )} over {3 "." "50"" s"} } } {} #" "= { {"120 J "+"1764 J"} over {3 "." "50"" s"} } {} # " "="538 W" {}} } {}

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     ( 1 hp = 746 W ) size 12{ \( 1" hp"="746"" W" \) } {} ! 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.

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Source:  OpenStax, College physics -- hlca 1104. OpenStax CNX. May 18, 2013 Download for free at http://legacy.cnx.org/content/col11525/1.1
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