If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Work-Energy Principle Example

The Work-Energy Principle tells us the amount of work done on an object or system will equal the change in kinetic energy for that object or system. Follow along in this worked example and understand about how to calculate the net work on an object and solve for the change in velocity. Created by David SantoPietro.

Want to join the conversation?

  • blobby green style avatar for user Wantedsolo
    Just had a half-stupid question that I thought I might as well ask...
    at In the last step of the video, David gets
    and equates that to

    So shouldn't the answer also equal -9 m/s? If so then how should we conceptualize this? Because anyways all our net forces/ velocities are in the same direction (left) which is considered +ve.

    Also does this have anything to do with Work being a scalar and that -ve work is only a relative quantity?

    Pls enlighten this dumb person...
    (5 votes)
    Default Khan Academy avatar avatar for user
    • leafers seed style avatar for user riyun3147
      In the ideal world of Math, you're right, technically taking the square root would give you both a positive answer, and a negative answer.

      However, in the case of Physics, as you also said, our velocity should be positive because this velocity has the same direction as the distance we've moved in.

      If you went for a hike up a trail, trying to reach a mountain, going back down would be considered negative, as you're losing your hard-earned distance up. It's opposite to the distance you've covered.

      If you're sledding down a hill, stopping your sled and going back up would be considered negative because you're going opposite to the distance you've been going.

      I like to consider up positive and down negative, but whether or not your velocity is negative or positive is relative to your distance. It depends on if the velocity is going with your distance (+), or against your distance(-).

      A velocity is positive (+) if that velocity is going in the same direction as our distance (d). ("Working with the distance is a positive thing to do!")

      A velocity is negative (-) when that velocity is going in the opposite direction of our distance (d). ("Working against the distance is a very negative thing to do. >:/ ")

      Looking at the problem, since there is a net force going left, and the distance we're going in is left, we know we should be speeding up. So our final answer of 9 would make more sense to be positive, going with our distance, as opposed to negative, which would mean we've started going backwards somehow.

      Hope this helps!
      (4 votes)
  • aqualine ultimate style avatar for user Simum
    What does net mean in net force?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • aqualine tree style avatar for user William 😊
    For those of you who are wondering at why V_f is POSITIVE 9 and not -9:

    Note that the problem asked for the final speed not velocity. Remember back in the first lesson, Sal explained Speed is a Scaler Quantity, not a Vector Quantity like velocity. A Scaler Quantity only has a magnitude, no direction, so it's always NON-NEGATIVE.
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Bing Li
    A 1,550 kg roller coaster cart is pulled at a constant speed of 4.3 m/s to the top of a
    53.5 m tall ramp that has a 45 degrees incline. The coefficient of friction between the ramp and the roller coaster cart is μ=0.052. When the roller coaster cart makes it to the top of the ramp, the cable suddenly breaks and the roller coaster cart accelerates back down to the bottom of the ramp. How much work is done by friction?
    I can understand the explanations for this practice question, but I don’t know what went wrong with my thinking:
    When the cart is at the top, it has potential energy (KE=0), which is the transformed to KE as it rolls down the ramp. Therefore, we can use the work/energy principle. Epotential=change in KE=Wg + Wf. KE=mgh Wg=Fg*(square root of 2) / 2 h(square root of 2)=mgh. Then Wf = 0.
    Intuitively this result doesn’t make sense when I know there is friction applied to the length of the ramp, but what is wrong with my thinking, please?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • starky sapling style avatar for user blackberry ❤
      Note that there is friction applied throughout the length of the ramp with coefficient μ=0.052. This problem does not test the work-energy principle, and relies on the formula for work itself:

      W_f = -fs = -(μmg cos(θ))*(2h/sin(θ)).

      This formula will be covered later on in the course :)
      (1 vote)

Video transcript

- [Instructor] So, the work-energy principle states that the net work done on an object is gonna equal the changing kinetic energy of that object. And this works for systems as well. So, the net work done on a system of objects is gonna equal the change in the total kinetic energy of the objects in that system. Now, that sounds really complicated and technical, but I like to think about the work-energy principle's a shortcut. This is a really nice shortcut that lets me determine the change in kinetic energy without having to do a bunch of complicated conservation of energy equations or kinematic formulas. The catch is that I need to know how to figure out what the net work is. So, how do you figure out net work? Well, the formula for work done is F d cosine theta. Since this formula, work-energy principle, relies on net work, this has to be magnitude of the net force times magnitude of the distance traveled times cosine of theta. Remember this theta has to be angle between, not any angle, angle between the net force direction and the direction of motion. And so, let's try this out. How do you use this thing? Let's kick the tires. Let's say there's a satellite. It's moving to the right and there's a net force on this satellite. Now, this net force could go in any direction. If the net force has a component in the direction of motion, then, the net work is gonna be positive. And if so, anything here from like negative 90, well, like it's negative 89.9, because 90 would be perpendicular for many, for like 89.9 negative to positive 89.9, you've got a component in the direction of motion. That means you're gonna be doing positive net work. And that means the change of kinetic energy will be positive because it just equals that number. That means kinetic energy increases. You're gonna be speeding up And that kind of, it makes sense intuitively. If your force is in the direction of motion, you're speeding up. What about the other case? What if your net force points in the opposite direction of motion? Well, now, the net work is gonna be negative. You'll have a negative change in kinetic energy. In other words, you're gonna slow down and if the net force points perpendicular, well, then, you're not doing any work because cosine of 90 is gonna be zero. No net work would be done. There's gonna be no change in kinetic energy. That doesn't mean you stop. It just means you're not going to speed up or slow down. This does something. You might be like, "Don't you do something?" Yeah, you're gonna drift upward. You're gonna start changing your direction, but this is not gonna be doing any work on you at that moment. And so, just to be clear, I mean, let's just try a complicated one here. Let's say this force goes in some direction. Let's say your velocity even goes down. So, maybe your satellite's heading downward and your force is gonna go in any direction. Well, if it goes this way, exactly backwards, it's gonna be 180. You're gonna be doing negative work. You're gonna be slowing down, decreasing kinetic energy. And you're not gonna change direction. If you're like this, you have a component opposite. So, you're gonna be slowing down and changing direction. This will just be changing direction. You're not speeding up or slowing down at that moment. This will speed you up and change your direction. And finally, this will just be speeding you up and you will not be changing your direction. So, the work-energy principle's convenient to just get a conceptual or qualitative idea of what's going on. And it can obviously also give you an idea of how to calculate things. So, let's try one where you actually have to get a number. So, let's say there's a hot air balloon and it's a 300 kilogram hot air balloon. Drifting to the left, it had an initial speed of seven meters per second, and it's traveling a total of 50 meters to the left during this journey. Now, there's gonna be forces on this hot air balloon. Obviously, there's gonna be gravity and some buoyant force, but because these are perpendicular to the direction of motion, they do no work. And so, when we're gonna use this work-energy principle, they're not even gonna factor in. We don't even have to know these since they were perpendicular and did no work. You only consider the forces in the direction of motion. So, let's say there was a wind gust helping you to the left here of 200 newtons, but this is a big, bulky balloon, not that aerodynamic. And so, there was a drag force from air resistance of 104 newtons to the right. And what we wanna know is we wanna determine the final speed of the hot air balloon after it travels 50 meters directly to the left with the forces shown. Now, there's lots of different ways to do this. You know, Newton's laws, you can do a kinematic formula, there's all kinds of stuff, even momentum, technically, impulse, but the easiest, I'm pretty sure the easiest way to do this is just gonna be the work-energy principle, which states that the net work done is gonna equal the change in kinetic energy. So, let's go ahead and do it. So, we know that net work is equal to the magnitude of the net force times the magnitude of the distance traveled times cosine of the angle between them. What's change in kinetic energy mean? Well, change in anything is final minus initial. So, since kinetic energy is 1/2 m v squared, this is just gonna be 1/2 m v final squared, minus 1/2 m v initial squared. So, that's gonna be the change in kinetic energy, final minus initial. All right, let's plug in numbers here. So, net force, how do we get that? Vertical pieces don't matter. We're just looking horizontally here. Those were the only ones that are gonna affect it. These vertical ones just cancel. So, we have 200 to the left, 104 to the right. So, we're gonna have to subtract those to get 96 to the left. And we just want magnitude. So, I'm gonna get 96 newtons to the left, not negative or anything, I'm just taking magnitude, times the distance traveled. We know that's 50. So, we get times 50 meters, cosine of the angle. Now, this is careful. You might be like, "Oh 180 here." But no, the net force points to the left. This 200 is winning here. So, leftward net force and the leftward direction, the angle between those two is zero, cosine of zero is just one. We're maxed out here. So, net force points in the direction of motion. So, equals, let's plug in the rest of this, 1/2, the mass is 300 kilograms, times v final squared, that's what we wanna determine, minus 1/2, 300 kilograms, times the initial speed is seven meters per second. So, we got seven meters per squared. Well, 96 times 50 is gonna come out to 4,800. That means that's the net work done. Notice that's joules. That's how much energy we've added. That's the change in kinetic energy. So, we know the net work is change in kinetic energy. We've added 4,800 joules of kinetic energy. That's gonna have to equal, half of 300 is 150 kilograms, times v final squared, minus, if you take a half of 300 times seven squared, you're gonna get 7,350. So, this is how much energy the hot air balloon started with initially. So, after we moved that to the left, we add those together. I'm gonna get 12,150 joules is how much kinetic energy the balloon ends with. And that's gotta equal 150 kilograms times v final squared. I could divide 12,150 by 150 and you get exactly 81 and that's gonna equal v final squared. And if you take a square root of that, you get exactly nine. So, the final velocity of this hot air balloon is nine meters per second. It sped up. That's not surprising. This net force was directed leftward and the object was moving leftward. So, we're doing positive work. We're increasing the kinetic energy. We started with seven meters per second. We ended up with nine meters per second. And this is an example of how you use the work-energy principle. So, to recap, the work-energy principle states, the net work is equal to the change in kinetic energy. This can help you just conceptually or qualitatively determine whether something is gonna speed up, slow down, or change direction or both. And then, quantitatively, you can use this to specifically solve for the change in kinetic energy, as well as the final or initial speed something might've had.