Power - Online Tutor, Practice Problems & Exam Prep

Power is defined as the rate at which work is done or energy is transferred, expressed mathematically as $P=\frac{W}{\mathrm{\Delta t}}$ or $P=\frac{E}{T}$, where P is power, W is work, E is energy, and T is time. The unit of power is the watt (W), equivalent to joules per second (J/s). Understanding power is crucial for calculating energy consumption, such as in a 100 W light bulb using 360,000 joules in one hour, or determining the average power delivered by a car engine during acceleration.

Power

Guys, up until now, all of our problems have involved work and energy. In some problems, you'll need to calculate or find how quickly that energy gets transported or is done to another object. And that's exactly what power is. So, in this video, I'm just going to quickly introduce you to the equation for power. We're going to see what it is, and we're going to check out some problems. The equation for power states that the average power is equal to the amount of work done by an object divided by the change in time. It measures how quickly that work gets transferred to something else. This is often called average power because we calculate it between two points in time. Another way we can write this is by using the relationship between work and energy, as work is just a transfer of energy. Sometimes, you might actually see this equation written as ET. The unit we use for power is called watt, symbolized by 'W', which can be confusing. So, don't confuse this 'W' with the work done. Unfortunately, they both use the same letter, but a watt is really just a joule per second, representing energy divided by time.

We've actually seen this diagram before, which relates all of the important variables we've seen so far in the chapter like forces, works, and energies. Now, we are just adding another branch, another connection between work and power. As mentioned, power is about how quickly we are doing work, expressed as WtΔ. Let's go ahead and check out some examples.

Let's calculate how many joules of energy are used by a 100-watt light bulb. We have the power, P = 100, and we need to find out how much energy it uses in an hour. This is in hours, and we want it in seconds, so this will be 60 \text{ minutes} \times 60 \text{ seconds per minute}, which equals 3600 seconds. We have our average power equation, which is equal to the work done divided by time, and it's also related to the amount of energy consumed per time. We're trying to solve for 'E', which is E = P \cdot t, resulting in 100 \text{ watts} \times 3600 \text{ seconds} = 360,000 \text{ joules}. Most of your problems will be solved using this straightforward equation.

Moving on, let's consider moving objects, exerting forces, and changing velocities. Here we have a 1300 kg sports car on flat ground. It starts from rest, meaning that the initial velocity is 0. It is accelerating due to the engine's force, labeled F_engine. At a later time, the car's velocity reaches 40 m/s over a period Δt = 7 seconds. We need to figure out the average power delivered by the engine. This P_average is calculated because the force of the engine moves the car through some distance, doing work in the process. To calculate this power, we consider the work done divided by the change in time or the energy divided by time. We know that work has been done, and with the change in time known as 7 seconds, we can determine the kinetic energy change: ΔKE = 12mvfinal2 - 12mvinitial2. The kinetic energy change lets us plug in this expression for our work to find the average power, leading to P_average = 12mv2Δt, yielding 1.49 \times 10^5watts = 149 \text{ kilowatts}. You can convert this to horsepower if needed. This is how much power is delivered by the engine during this acceleration.

Hopefully, that made sense. Thank you for watching, and I'll see you in the next video.

How much power must an elevator motor supply in order to lift a 1000 kg elevator at constant speed a height of 100m in 50 seconds?

• A. 100,000 W
• B. 0 W
• C. 2,000 W
• D. 19,600 W

Power of Pushing a Box

Hey everyone, let's check this out here. So we have a 15 kilogram box that is sliding across a frictionless surface, and its speed is going to increase. We're told that the initial speed of this block is 10 meters per second, but it's accelerating over some distance, and then at some point over here, its final velocity is going to be 30. We're told that the acceleration along this interval here, this "a", is equal to 2. That's basically all we know, and in the first problem, we want to calculate what is the change in mechanical energy of the block's mechanical energy. So the variable that we are looking for is Δme. That's the change in mechanical energy. Now remember, for a block's mechanical energy, it could be potential or kinetic. So really what they're asking for here is what is the total change in the potential energy plus the change in the mechanical energy. Alright. So let's consider each one of these terms here. Now we have a block that's not attached to any springs; there are no springs in this problem, and there's also no gravity because we're just going along the horizontal surface. So there's no change in the potential energy. So really all that happens here is that the block's kinetic energy is changing across this interval, and because it's going faster right from 10 to 30, there's a change in the mechanical energy. So that's Δme. This is just going to equal k_final−k_initial. So Δme here is just going to equal the formula for kinetic energy is 12mvfinal²−12mvinitial². Now I have all of those values. I have the masses and the velocities, so I can actually figure this out. My Δme is just going to be 12×15×30²−12×15×10². When you go ahead and plug this all into your calculator, you're going to get 6,000 joules. So one way you can also think about this here is that there's a force that is pushing this thing, this pushing this block that's why it's causing it to accelerate. We don't know what that force is, but that force is doing work on the object, and that work is equal to the change in the kinetic energy. So that work that you're inputting changes the mechanical energy by 6,000 joules. Alright? And that's one way you can think about it as well. Let's move on to the second problem. And the second problem we want to figure out, the rate at which energy is transferred. So that's a dead giveaway that this is actually talking about power. Remember that power is equal to the change in energy over the change in time. And because we're not told because we're basically told 2 points, right, from initial to final, we don't know what's happening in between; this is an average power. So all we really have to do here is realize that the change in energy is actually just the change in the mechanical energy that we just calculated. So if we want to figure out the power, all we have to do is just have the mechanical energy, which we already have, but we also need to know the time. We don't have anything and any information about time, so that's what we need. Actually, I'm sorry. This should be a question mark. We don't have anything about time, so we're going to have to go solve that in order to get power. So let's go ahead and do that. Right? So if we want time here, let's see. We're also told that the mass of this block, the information we're told is also the initial and velocity, the acceleration, and the final velocity. These are kinematics variables. So if I want to figure out t, which is also a kinematics variable, I just need to do exactly what I did when we talked about 1-dimensional motion. I need to set up my 5 variables and I need 3 out of 5. So I don't know t, that's actually what I'm going to be looking for here. So I have, vnaught, which is 10, vfinal, which is 30, and I have the acceleration which equals 2. The last one is Δx. It's the distance over which this thing is accelerated. I actually don't know what that is. All I'm told is the initial and final velocities and the acceleration in between; I don't know the distance. But luckily, I actually have 3 out of 5 variables. I have 1, 2, and 3. So the equation that's going to relate them and also give me time is going to be the first kinematic equation. This is basically just vfinal=vinitial+at. Change of velocity equals acceleration times time. So you rearrange for this because we really want to solve for this time over here, and you'll find that we're going to get vfinal−vinitiala. So we go and plug this in. This is going to be 30 minus 10 divided by 2, and this is going to give me 10, and this is going to be in seconds. So this number here, this 10 seconds, is now what I just plug into the power average equation, and I'm done. So this p_average is just going to be the change in mechanical energy, which is 6,000 joules divided by 10 seconds, and so that's going to give me an average power of 600 watts. Alright. So hopefully that makes sense, guys. Let me know if you have any questions. That's it for this one.

Power of a Winch on an Incline

Hey, guys. So let's check this one out together here. We have a block that is 1400 kilograms that gets pulled up by a winch and cable, which is basically just like a cable that's attached to a spinning or cranking motor. It's going to pull this block up the incline, so this would probably happen in like a rock quarry or something like that. I don't know. So, we're going to call this force of the cable the tension, right, that's tension by a cable. What I want to ultimately figure out is the power that is due to the force of this cable. So how do we figure this out here? Remember that the power equation is just ΔE over Δt, but it's also equal to the work that is done divided by Δt. Remember that power - sorry, that work and change in energy mean the same thing. So what we're actually going to do here is because we're trying to figure out the force of the cable, we're actually going to use this work equation. Now what I have is I actually have the Δt and that's just going to be the 25 seconds. That's this Δt over here. And remember that work is always equal to force times distance. So what I can do here is I can set this up as the tension force times the distance that this block gets pulled up, I'm going to call this some d here, times the cosine of the angle between those two vectors. Right? So this t and this d. They both point in the same direction up the incline, therefore, the angle between them is 0 and this just equals 1. And now we're just going to divide this by Δt. So in order to figure out the power, I'm gonna actually need to figure out the tension force. I'm gonna actually have a number for it and I also need to figure out the distance that this block gets pulled up. I don't have either of those numbers. So how do we do this? First, I'm actually going to go ahead and list all the things I know about this problem. I already have the mass, I know that the incline angle is 37 degrees, and I know that this block is pulled up at constant speed. What that means is that the acceleration is equal to 0. I'm also told what the coefficient of kinetic friction is, μ_k is 0.4, and I have that this Δt here is 25 seconds. The only other thing that I know here is that the height of the incline, which is this over here not our d. This is I'm gonna call this h is equal to 60.2 meters. So let's go ahead and get started here. The first thing I want to do is figure out the tension that's in the cable. And because I know a couple of things about the forces in this problem, this is really just going to turn into a forces on an inclined plane problem. So remember that there's this force that's pulling this up, but the reason that this is pulled at constant velocity is because I have an mg, but this mg has a component down the incline, this is mg_x. And I also have a friction force, so this is going to be my friction _k that is preventing this thing, right? It's actually exerting a force downwards again because this thing is being pulled up at constant speed this way. So this is going to be my V. So if I go ahead and sort up in Newton's laws, I need to set up an F = ma in order to figure out this tension here. So, I'm going to have to do that. So F = ma. Now remember, what we know is that the acceleration is equal to 0, it's a constant speed, so we actually know that this is just going to be 0. So this allows me to set up an equation relating all my forces. I'm going to pick the upward direction to be positive because that's where the block is going. So therefore my tension is going to be positive, and this is going to be minus mg_x minus friction kinetic equals 0. When you move both of these terms over to the other side, they both become positive, and I'm just going to go ahead and expand them. Remember that mg_x is just equal to mg times the sine of θ, and remember that friction is on an inclined plane. It's μ_k times the normal force and the normal force is mg times the cosine of θ. We've seen this written a bunch of times, so hopefully that's familiar to you. So if I'm sorry, this is actually supposed to be positive as well. Now if you go through these variables, we actually have all of them. We have mass, we have g, we have the θ, we also have the coefficient. So I'm just going to have to plug this in. Unfortunately, this is going to be really long, but you can just plug it into your calculator and follow it along. So the mass is going to be 1400, g is 9.8, then we have the sine of 37 plus 0.4 times 1400 times 9.8 times the cosine of 37, just make sure your calculator is in degrees mode, and what you should get when you plug both of these things in and sort of add them together or you can just plug it in as long as expression is 12640 newtons. So that's the first variable I need. I need the tension. So that's done. The next thing I need is I need to figure out what the distance is along the cable because I don't have what that d value is. So let's go ahead and do that. What I'm going to do here is I'm going to draw a simplified version of the triangle because there's a lot going on here. So what's happening here is I've got this triangle like this and I've got some of the values. I've got the d, which is what I actually need, what I need, I need that distance and I also know what the angle of the incline is, 37 degrees, and I'm also told what h is, 60.2. Now remember for triangles, as long as I have one angle and one side, I can figure out any other piece of the triangle. So if I want to relate to the hypotenuse, which is d, and then the height, which is the opposite side, I'm going to have to use a sine function. I'm a sine function here. So remember that the sine of the angle is related to the opposite side which is h over the hypotenuse. So if I want to figure out what this, with what this distance is, and all I have to do is just trade these two these two variables and switch their places. So, d is equal to h over the sine of θ. So this equals the height which is 60.2 divided by the sine of 37. If you go ahead and work this out, what you're going to get is a distance of exactly 100 meters. So that's how far that this block gets pulled up the incline. So, this d here equals 100. And now we have everything we need to solve this problem here. So we're done with t and d. Now we just plug this all in. So the power is just going to be 12640 times the distance, which is 100, and we're going to divide it by the 25 seconds it takes. And when you go ahead and do that, you're going to get exactly 50,560 watts. That's how much energy this is, this has to this cable actually has to exert, this motor has to exert in order to pull this block up the incline like this. It's a lot of power. So, you know, you might see this written as 50.5 kilowatts. Right? That's another way you might see that written. Alright, guys. So that's it for this one.

Here’s what students ask on this topic:

What is the definition of power in physics?

Power in physics is defined as the rate at which work is done or energy is transferred. Mathematically, it is expressed as $P=\frac{W}{\mathrm{\Delta t}}$ or $P=\frac{E}{T}$, where P is power, W is work, E is energy, and T is time. The unit of power is the watt (W), which is equivalent to joules per second (J/s).

How do you calculate the power consumed by a 100 W light bulb in one hour?

To calculate the power consumed by a 100 W light bulb in one hour, you use the formula $E=PT$. Here, P is 100 W and T is 1 hour, which is 3600 seconds. So, $E=100\times 3600=360000$ joules. Therefore, the light bulb consumes 360,000 joules of energy in one hour.

What is the unit of power and how is it related to energy and time?

The unit of power is the watt (W), which is equivalent to joules per second (J/s). Power is related to energy and time through the formula $P=\frac{E}{T}$, where P is power, E is energy, and T is time. This means that power measures how quickly energy is transferred or work is done over a period of time.

How do you calculate the average power delivered by a car engine during acceleration?

To calculate the average power delivered by a car engine during acceleration, you can use the formula $P=\frac{W}{\mathrm{\Delta t}}$. First, find the work done using the kinetic energy theorem: $W=\frac{1}{2}m{v}^2$. Then, divide by the time interval Δt. For example, if a 1300 kg car accelerates from 0 to 40 m/s in 7 seconds, the work done is $\frac{1}{2}1300\times {40}^2$, and the average power is $\frac{1}{2}1300\times {40}^2÷7$, which equals 149 kW.

What is the relationship between work, energy, and power?

Work, energy, and power are closely related concepts in physics. Work is the transfer of energy through force applied over a distance, expressed as $W=Fdcos\theta$. Energy is the capacity to do work. Power is the rate at which work is done or energy is transferred, given by $P=\frac{W}{\mathrm{\Delta t}}$ or $P=\frac{E}{T}$. Essentially, power quantifies how quickly work is performed or energy is used over time.