Video Transcript
In this video, we’re going to learn
about power. We’ll learn how power is defined in
physics. And we’ll get some practice using
it practically.
To start out, imagine that, after
gym class, you and some friends decide to have a rope-climbing competition. Each of you grabs onto a rope and
at a signal starts to climb as fast as possible toward the ceiling. As time passes, the group begins to
separate out, with the faster climbers moving ahead. If all of you have about the same
weight, what is the physical factor that separates the faster climbers from the
slower ones? To better understand this question,
we’ll want to learn a bit about power.
The term power is a little bit like
the term work in the sense that we hear it often. But in physics, this term has a
very specific meaning. Power is defined as the rate of
energy used or work done over time. We can symbolize power with a
capital 𝑃. And it’s equal to energy per unit
time or work done in the amount of time it took to do that work.
Think of the example of climbing up
a flight of steps. If you have a mass 𝑚 and the
height of the stairway is ℎ, then the work you perform to get to the top is equal to
the force of gravity on you, 𝑚 times 𝑔, multiplied by the distance you go against
that force ℎ. Let’s imagine that you climb the
stairs very slowly that you took 60 seconds to go from the bottom to the top. Then let’s say you start over and
climb the steps this time as fast as you possibly can. On the second run through, you only
take five seconds. And you do the same amount of work
by moving the same amount of mass through the same height. By the definition of power, this
means that the power you used to climb the stairs the first time and the power you
used to climb the stairs the second time is different. In fact, the power the second time
is 12 times as great as the power from the first time.
So we see that power is especially
useful as a way of measuring the rate of energy usage or work done. Power is measured in units called
watts, named after the inventor James Watt, where one watt is defined as a joule per
second. Recall that joule is the unit we
measure both energy and work in. When we calculate or solve for
power, we can consider power over a certain time span, that is, average power, or
the power at a particular instant in time, instantaneous power. To calculate average power, we use
a time interval Δ𝑡. Whereas for instantaneous power, we
shrink that time interval down until it’s infinitesimally small. Instantaneous power is essentially
a time derivative of energy. Let’s get some practice with this
idea of power in physics through a couple of examples.
A cyclist in a race must climb a
hill sloping at 7.5 degrees above the horizontal. The cyclist moves with a speed of
5.3 meters per second. The mass of the cyclist and their
bike is a total of 85 kilograms. What power output is required from
the cyclist?
We can call this power output 𝑃
and start off by drawing a diagram of this scenario. In this scenario, a cyclist climbs
a hill inclined at an angle we’ve called 𝜃 of 7.5 degrees above the horizontal. The cyclist moves along the ground
at a speed 𝑣 of 5.3 seconds and along with their bike has a mass 𝑚 of 85
kilograms. We can recall that power is equal
to work done over the time it takes to do that work. In this case, the cyclist is doing
work against gravity by moving uphill. Recalling also that work is equal
to force done multiplied by distance traveled. If ℎ is the height that the cyclist
ascends against gravity, then we can say that the work done is equal to the force of
gravity, the cyclist mass times 𝑔, multiplied by ℎ, where 𝑔 is a constant value of
9.8 meters per second squared.
Writing out our equation for power
𝑃, work over time, we can say that our time interval is one second. That means ℎ is the vertical
distance the cyclist ascends in one second. Because we know the cyclist’s speed
is 5.3 meters per second, that means our right triangle we’ll use to solve for ℎ has
a hypotenuse of 5.3 meters. We can write ℎ, therefore, as 5.3
meters times the sin of 𝜃, where 𝜃 is 7.5 degrees. Plugging this expression in for ℎ
in our equation for power, we now have an expression for power where all the
variables are known or can be calculated.
Plugging in for 𝑚 and 𝑔, when we
enter this expression on our calculator, to two significant figures, we find it’s
580 watts. That’s the power output needed from
the cyclist to climb this hill, maintaining a speed of 5.3 meters per second.
Let’s look at a second example
involving power.
A 55.0-kilogram woman in a gym does
50 deep knee bends in 3.00 minutes. In each knee bend, her center of
mass is lowered and raised by 0.400 meters. She does work in both
directions. Assume her efficiency is 20
percent. Calculate the energy used to do all
50 repetitions in units of kilojoules. What is her average power
consumption in watts?
In this two-part exercise, we want
to solve for the energy — we can call it capital 𝐸 — and the average power
consumption — we can call it capital 𝑃 — involved in this woman exercising. Let’s start with a diagram. In this scenario, a woman starts
out upright and in doing a deep knee bend lowers her center of gravity at distance ℎ
given as 0.400 meters. She then stands back up and raises
her center of gravity that same vertical distance. We’re told the woman does 50 of
these deep knee bends in a time 𝑡 of 3.00 minutes and has an efficiency we’ve
called lowercase 𝑒 of 20 percent or 0.20.
We wanna start by solving for the
total energy required for her to go through 50 of these motions. We can say that the total energy
output by the woman is equal to the total work she does divided by the efficiency of
that work against gravity. Recalling that work is equal to
force times distance, we can write that the total work 𝑤 is equal to the force of
gravity on the woman, 𝑚 times 𝑔, multiplied by the distance she moves through
gravity, ℎ, all multiplied by two times the number of repetitions she does. We’ll have that factor of two
because she moves down and then up.
Treating the acceleration due to
gravity as exactly 9.8 meters per second squared, we see we’ve been given the
efficiency 𝑒, mass 𝑚, we know 𝑔, given ℎ, and we know capital 𝑁. So we’re ready to plug in and solve
for 𝐸. Plugging in the values for the
efficiency 𝑒, mass, 𝑔, ℎ, and two times 𝑁, when we calculate the total energy
used 𝐸, we find it’s equal to 108 kilojoules. That’s the total amount of energy
the woman would need to supply to do all these exercises.
Next, we want to solve for the
power output of the woman over the course of these exercises. Recalling that power is equal to
energy per unit time, we can write that 𝑃 is equal to 𝐸, the result we solved for
in part one, divided by 𝑡, the time value given as 3.00 minutes. When we plug in our two values,
we’re careful to convert our time into units of seconds. That’s because power is given in
units of joules per second or watts. Calculating this fraction, to three
significant figures, it’s 599 watts. That’s the power the woman would
need to supply to do these exercises in this amount of time.
Let’s summarize what we’ve learned
so far about power.
We’ve seen that power is the rate
at which work is done or energy is used. In general, power is equal to
energy used per unit time. We’ve also seen that power can be
measured as an average, the change in energy over some finite change in time, or an
instantaneous power can be measured, which is the time derivative of energy. And finally, we’ve seen that power
is measured in units called watts, where one watt is defined as a joule of energy
used up in one second.
