Lesson Video: Power

In this video we learn the physics definition of power and how to calculate average and instantaneous power in watts from work done or energy used in a certain time.


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.

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