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Spring potential energy example (mistake in math)

A spring, a frozen loop-d-loop and more! Explore the concept of energy conservation with a thrilling ice loop-d-loop scenario as you see how potential energy from a compressed spring can propel a block of ice through a loop, using kinetic and gravitational energy. (See if you can find the mistake I made and get the right answer!). Created by Sal Khan.

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  • male robot johnny style avatar for user sahildahake
    Can anyone explain to me what is the mistake in sal's calculations?
    Is it taking potential energy as 10x^2 instead of 5x^2, or is it something else?
    (75 votes)
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  • blobby green style avatar for user Gabriel Os
    I'm sorry but according to the diagrams it seems as if the cube would be upside down when it reaches the bottom side of the curved surface. Wouldn't that mean that the the downward force wouldn't just be the force of gravity but also the normal force? I do believe the only way to have only a downward gravitational force would be if the ice is on the top side of the curved surface, but the diagrams makes it seem otherwise the opposite.
    (14 votes)
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    • leaf green style avatar for user Trisa
      By definition, the normal force is the force perpendicular to the surface of the plane an object contacts. In other words, it is the force exerted by the surface of a plane that resists an object going through it (remember, Newton's law about equal opposite forces). (Think about your hand pressing into a table as you lean on it. If the normal force isn't enough to resist the force of your hand, your hand will go right through it.)

      In most cases, when an object rests on top of a surface--such as a block on the ground--the normal force points upward while the gravitational force points downward. However, this isn't always the case (these forces do not always point completely opposite each other)--the definitions hold but the direction changes with the position of the object and surface of the plane it rests on. For example, when a block is on a inclined plane, the gravitational force is still downward (or toward the center of the earth) and the normal force is still perpendicular to the surface of the plane, but it is at angle (leaning away from the vertical axis the gravitational force is on) from its original position when it was resting on the ground.

      So at the top of the loop de loop, the normal force is downward,but it only works to resist the ice block breaking the through the surface and leaving the loop (otherwise, the ice block might fly out of the loop), but not to push the block in a particular direction. Likely, because gravitational force is downward (or away from the surface of the icy loop), there may be little or no normal force at the top of the loop since there is no noticeable force pushing against the loop's surface. I believe this is why the normal force can be ignored in these calculations. Again, gravitational force is always toward the center of earth (that is, if your problem scenario is on the earth), which is still downward.

      Hope this helps. :)
      (39 votes)
  • leafers tree style avatar for user mmvasquez12
    If Sal didn't do the mistake, is "x" suppose to have come out to about 3.95m?
    (6 votes)
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    • blobby green style avatar for user avanna
      I got x to equal about 4.5m because I rounded g to 10m/s^2. The mistake is in the equation of PE. PE=1/2 K X^2. Since K is equal to 10 in the example he used, PE=(1/2) (10) x^2 which simplifies to PE=5X^2, If you continue this problem and solve for x..... 1/2mv^2 + mgh=5X^2 you should get about 4.5m. Sal forgot about the 1/2 and continued the problem with PE=10X^2 instead of 5X^2. Hope that helps :)
      (24 votes)
  • blobby green style avatar for user 😊
    doesnt the spring have kineic energy also when it is moving horizontally?
    why only potential energy?
    (4 votes)
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  • stelly green style avatar for user Vedant Karandikar
    Why do we not use Normal force for centripetal force they both are in the same direction and also has magnitude(in this case 4x10=40 Newton)?
    (3 votes)
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    • male robot hal style avatar for user Andrew M
      Normal force provides part of the centripetal force, gravity provides the rest
      If you are trying to find the minimum possible speed, that's the speed where gravity provides all the acceleration at the top, and the normal force provides none
      (7 votes)
  • male robot hal style avatar for user Rafiuddin Galib
    I am still confused as to why the centripetal acceleration is the acceleration of gravity.
    In the video, for the net force, he includes the force of gravity and the normal force which points towards the center and equaled it to v^2/r which is the centripetal acceleration
    (5 votes)
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    • male robot hal style avatar for user Andrew M
      If the ball is going to just make it through the top of the loop, then that means the ball is right on the edge of losing contact with the track at the top. That means the track is not applying force to the ball. That means the only force available at that moment to provide the necessary centripetal acceleration at that moment is the weight of the ball. If the ball is going very fast then the track will also have to provide some of the necessary centripetal force at that moment; the weight of the ball won't be enough.
      (3 votes)
  • blobby green style avatar for user akshuagarwal12
    Can you please tell me that why the centripetal acceleration =9.8m/s^2 ?
    (4 votes)
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  • purple pi purple style avatar for user Pranay Narayankar
    What does the value of spring constant " K" depend upon?
    thank you
    (4 votes)
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  • piceratops ultimate style avatar for user Hari
    Isn't the answer sqrt(2g) and not sqrt(g)?
    (4 votes)
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  • leaf green style avatar for user Bayan Atawna
    at , why did you chose that point ( the top of the loop de loop) to be the Efinal, why didn't you chose the bottom of the loop de loop??
    (3 votes)
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    • orange juice squid orange style avatar for user Hanry Hu
      The energy is always the same in the system (Law of Conservation of Energy), so the Estart = Efinal = Emiddle -- this means that you can choose any point to try to help you find the total energy or the potential energy of the spring. However, if you tried to find the E at a spot on the side or bottom of the loop, the calculation would be harder because the centripetal acceleration (towards the center of the loop-de-loop) would not be the gravitational acceleration, so there is not enough information to find the velocity to find the Kinetic Energy.

      By choosing a spot on the top of the loop-de-loop, it gives you an extra known variable during your calculation, the "centripetal acceleration," because the acceleration towards the center of the loop-de-loop coincides with the (gravitational) acceleration towards the ground.
      (3 votes)

Video transcript

Welcome back. So let's do a potential energy problem with a compressed spring. So let's make this an interesting problem. Let's say I have a loop-d-loop. A loop-d-loop made out of ice. And I made it out of ice so that we don't have friction. Let me draw my loop-d-loop. There's the loop, there's the d-loop. All right. And let's say this loop-d-loop has a radius of 1 meter. Let's say this is-- this right here-- is 1 meter. So of course the loop-d-loop is 2 meters high. And let's say I have a spring here-- it's a compressed spring. Let's say this is the wall. This is my spring, it's compressed, so it's all tight like that. And let's say its spring constant, k, is, I don't know, 10. Attached to that compressed spring-- so I have a block of ice, because I need ice on ice, so I have no friction. This is my block of ice, shining. And let's say the block of ice is, I don't know, 4 kilograms. And we also know that we are on Earth, and that's important, because this problem might have been different if we were on another planet. And my question to you is how much do we have to compress the spring-- so, let's say that the spring's natural state was here, right, if we didn't push on it. And now it's here. So what is this distance? How much do I have to compress this spring, in order for when I let go of the spring, the block goes with enough speed and enough energy, that it's able to complete the loop-d-loop, and reach safely to the other end? So, how do we do this problem? Well, in order-- any loop-d-loop problem, the hard part is completing the high point of the loop-d-loop, right? The hard part is making sure you have enough velocity at this point, so that you don't fall down. Your velocity has to offset the downward acceleraton, in which case-- and here, is going to be the centripetal acceleration, right? So that's one thing to think about. And you might say, wow this is complicated, I have a spring here, it's going to accelerate the block. And then the block's going to get here, and then it's going to decelerate, decelerate. This is probably where it's going to be at its slowest, then it's going to accelerate back here. It's a super complicated problem. And in physics, whenever you have a super complicated problem, it's probably because you are approaching it in a super complicated way, but there might be a simple way to do it. And that's using energy-- potential and kinetic energy. And what we learned when we learned about potential and kinetic energy, is that the total energy in the system doesn't change. It just gets converted from one form to another. So it goes from potential energy to kinetic energy, or to heat. And we assume that there's no heat, because there's no friction. So let's do this problem. So what we want to know is, how much do I have to compress this spring? So what I'm essentially saying is, how much potential energy do I have to start off with-- with this compressed spring-- in order to make it up here? So what's the potential energy? Let's say I compress the spring x meters. And in the last video, how much potential energy would I then have? Well, we learned that the potential energy of a compressed spring-- and I'll call this the initial potential energy-- the initial potential energy, with an i-- is equal to 1/2 kx squared. And we know what k is. I told you that the spring constant for the spring is 10. So my initial potential energy is going to be 1/2 times 10, times x squared. So what are all of the energy components here? Well, obviously, at this point, the block's going to have to be moving, in order to not fall down. So it's going to have some velocity, v. It's going tangential to the loop-d-loop. And it also is going to have some potential energy still. And where is that potential energy coming from? Well, it's going to come because it's up in the air. It's above the surface of the loop-d-loop. So it's going to have some gravitational potential energy, right? So at this point, we're going to have some kinetic energy. We'll call that-- well, I'll just call that kinetic energy final-- because this is while we care about alpha, maybe here it might be the kinetic energy final, but I'll just define this as kinetic energy final. And then plus the potential energy final. And that of course, has to add up to 10x squared. And this, of course, now, this was kind of called the spring potential energy, and now this is gravitational potential energy. So what's the energy at this point? Well, what's kinetic energy? Kinetic energy final is going to have to be 1/2 times the mass times the velocity squared, right? And then what's the potential energy at this point? It's gravitational potential energy, so it's the mass times gravity times this height. Right? So I'll write that here. Potential energy final is going to be mass times gravity times the height, which also stands for Mass General Hospital, anyway. You can tell my wife's a doctor, so my brain just-- anyway. So let's figure out the kinetic energy at this point. So what does the velocity have to be? Well, we have to figure out what the centripetal acceleration is, and then, given that, we can figure out the velocity. Because we know that the centripetal acceleration-- and I'll change colors for variety-- centripetal acceleration has to be the velocity squared, over the radius, right? Or we could say-- and what is the centripetal acceleration at this point? Well it's just the acceleration of gravity, 9.8 meters per second squared. So 9.8 meters per second squared is equal to v squared over r. And what's the radius of this loop-d-loop? Well it's 1. So v squared over r is just going to be equal to v squared. So v squared equals 9.8-- we could take the square root, or we could just substitute the 9.8 straight into this equation, right? So the kinetic energy final is going to be equal to 1/2 times the mass times 4 times v squared times 9.8. And that equals-- let's just use g for 9.8, because I think that might keep it interesting. So this is just g, right? So it's 2 times g. So the kinetic energy final is equal to 2g-- and g is normally kilogram meters per second squared, but now it's energy, right? So it's going to be in joules. But it's 2g, right? And what is the potential energy at this point? Well, it's the mass, which is 4, times g times the height, which is 2. So it's equal to 8g. Right. So what's the total energy at this point? The kinetic energy is 2g, the potential energy is 8g, so the total energy at this point is 10g. 10g total energy. So if the total energy at this point is 10g, and we didn't lose any energy to friction and heat, and all of that. So then the total energy at this point has also got to equal 10g. And at this point we have no kinetic energy, because this block hasn't started moving yet. So all the energy is a potential energy. So this also has to equal 10g. And this g, I keep saying, is just 9.8. I just wanted to do that just so you see that it's a multiple of 9.8, just for you to think about. So what do we have here? [? I'll do ?] these numbers worked out well. So let's divide both sides by 10. You get x squared is equal to g, which is 9.8. So the x is going to be equal to the square root of g, which is going to be equal to what? Let's see-- if I take 9.8, take the square root of it, it's like 3.13. So x is 3.13. So we just did a fairly-- what seemed to be a difficult problem, but it wasn't so bad. We just said that, well the energy in the beginning has to be the energy at any point in this, assuming that none of the energy is lost to heat. And so we just figured out that if we compress this spring, with the spring constant of 10. If we compress it 3.3 meters-- 3.13 meters-- we will have created enough potential energy-- and in this case, the potential energy is 10 times 9.8, so roughly 98 joules. 98 joules of potential energy to carry this object all the way with enough velocity at the top of the loop-d-loop to complete it, and then come back down safely. And so if we wanted to think about it, what's the kinetic energy at this point? Well we figured out it was 2 times g, so it's like 19.6 joules. Right. And then at this point, it is 98 joules. Right? Did I do that right? Well, anyway I'm running out of time, so I hope I did do that last part right. But I'll see you in the next video.