Spring potential energy is the store energy that drives the restoring force described in Hooke's Law. Other than Hooke's Law, the equation for the potential energy function, U=1/2kx^2, is essentially used when determining the spring potential energy.
So let's talk about Spring Potential Energy. Alright just to remind you whenever we've got a conservative force what happens is the work done by that force can be expressed as minus the change in a function that depends only on position and that's called the potential energy function U some times people use PE for potential energy but let's just use U.
Alright so work is always equal to force times change in position but in this case our force depends on position so I don't know what to use for my force if my position is changing which one do I use? Do I use kx at the beginning kx at the end kx half way through I don't know hmm so what we're going to do is we're going to make a graph of force versus position so the idea is that the force here depends on position but I want to somehow find a way to interpret force times change in position in this graph alright. Well the force is vertical the change in position if I just change the position a little bit is horizontal so if I do force times change in position then it gives me the area of this low red box and as long as the change in position is really really small the force won't change very much and then I can just consider it constant over that little box so that means that I can interpret work as the area under a force versus position graph this is always true whenever we have a conservative force. Now in our case the force is Ã¢Â€Â“kx so that means that the change of potential energy will be the area under kx. Alright here's kx Ã¢Â€Â“f and it just looks like that and I want to know what's the work done as I go from x initial to x final and the work is just the area under this graph but that's easy to calculate look at it it's a trapezoid well it's a trapezoid on its side oh well. Area of a trapezoid is very very simple it's equal to the height times the average of the two bases alright well the two bases are kx initial and kx final so in order to get that average I just add them together and divide by 2 and then I'm going to multiply by the height now remember it's a trapezoid on its side so by height I mean this distance x final minus x initial so there we go now I've got an overall factor here of k over 2 and once I take that one out it's a difference of two squares x final minus x initial times x final plus x initial well that's x final squared minus x initial squared so the change of potential energy is the change in one half kx squared so that means that the potential energy stored in a spring is just one half spring constant stretching squared there we go let's do some problems.
Alright so we're given that the energy stored in a spring is 20 joules when it's stretched 5 centimeters and we want to know how much energy is stored when it's stretched 15 centimeters. Alright, the straight forward way to solve this problem would be to just go ahead and use U equals one half kx squared plug in 20 joules plug in 5 centimeters really you'd have to plug in 0.05 [IB] we want to do it in S.I units solve for the spring constant and then just use the formula again but there's a much easier way to do this problem look 5 centimeters to 15 centimeters what did I do? I tripled the displacement I tripled the stretching well jeez U is one half kx squared so if I triple x what's 3 squared? It's 9 so that means that I'm going to 9 triple the potential energy well 9 times 20, 180 joules very clean very simple I don't need to solve for the spring constant nobody asked me for the spring constant so I don't want to waste time doing that so this way of solving these types of problems is a very very very useful thing to get used to get used to.
Alright so the next example we've got a 200 grams mass and I'm stretching its spring 40 centimeters with that 200 grams mass and I want to know how much energy is stored in the spring. Alright in this case I don't have the spring constant so I can't directly use this formula so what I'm going to do first is get the spring [IB] get the spring constant so I have force equals kx the force that's stretching it is point 2 times 10 remember you've got to multiply by G alright because its force equals kx not mass equals kx alright so we've got point 2 times 10 equals k times what is it? Point 4 alright and then we'll divide and we'll have k equals 2 over point 4 well what is 2 over point 4? Well it's 1 over point 2 which is five and what's the unit? Well k is Newtons per meter so Newtons per meter alright we're not done yet jeez it takes forever alright so we need to find out how much energy so we'll do one half k and then x point 4 squared right? Okay so this will be one half times point 4 squared point 4 is 4 times 10 to the minus 1, I square it I get 16 times 10 to the minus 2 over 2 and then times 5 right? 16 divided by 2 is 8, 8 times 5 is 40 so when we multiply we'll get 0.4 joules because 40 divided by 100 is point 4. Alright let's do this the easier way there's a reason I left some room over here.
In my experience, students would rather do it this way it takes a lot more time it's a lot more kind of messy but its more direct its more kind of oh I need the spring constant let me get the spring constant let me show you another way that we could do it that's much much much cleaner. Alright I don't have the spring constant, so what I'm I going to do? Well I solved through the spring constant over here but the idea is I don't really need the spring constant I want the energy so let's try to figure out what the energy is in terms of the weight and the displacement so we'll say one and a half the spring constant is mg divided by x and then we'll have x squared one of the x's cancels and so that gives me U is one half mgx watch how beautiful this is. One half point 2, 10, point 4 there we go. 10 times point 2 is 2, 2 divided by 2 cancel cancel cancel answer 0.4 joules very quick very very very simple but you have to kind of think Algebraically and I think that that's the issue that sometimes students have because they're not used to it. Work out a couple of problems like that, try to get used to this thinking algebraically because it makes stuff a lot easier. Anyway that's potential energy stored in springs.
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