Elastic Potential Energy of a Spring

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. You take a two kilogram block and lower it gently onto a spring, until it hangs at rest. The block loses two joules of gravitational potential energy. But the spring stores only one joule. Energy is always conserved, right? So where did the other joule go? Hmm, strange! By the end of this lesson, you will know. Spring energy questions come up often in JEE, so let's begin.

0:33Energy stored in a spring

So, look at the figure. On top, a spring joins a wall to a box, at its natural length; the dashed line marks x equals zero. Below, the box is pulled out by x, and the spring pulls back with force k x, the arrow. That is why the label below says restoring force, F equals minus k x: the minus sign means the force acts against the stretch. Now, you did work to stretch this spring. Where did that work go? Into the spring, yes! It is stored as elastic potential energy, which we call capital U, okay? U equals half k x squared: k is the spring constant, and the small two on x means squared. That is the bowl-shaped curve on the right, U against x. The label there gives the work to stretch it from x one to x two, x with a small one and a small two: half k times, x two squared minus x one squared. Careful: on this U graph, that work is how much the curve rises from x one to x two, not the shaded area under it. Area gives work only on a force graph. We met that in Chapter one. Clear? Now, careful with x. It is the extension or the compression from the natural length, not the full length. So tell me. Stretch the spring by some x, or compress it by the same x. Which one stores more energy? Neither! The same energy, both ways. x comes squared, so its sign does not matter at all. See the bowl in the graph: it rises the same on both sides of zero. Right? Now, the force that holds the spring stretched is F equals k x. Put x equals F over k into half k x squared, and you get F squared over two k. Or mix the two, and you get half F x. The double arrow means therefore. Three forms of one energy: use whichever fits the data. Got it? Quick check. You double the stretch x. The stored energy doubles too, right? No, no, careful! Four times! x is squared, and two squared is four. And if you double the force instead? F is squared too, so again four times. With me?

2:45A moving block runs into a spring

Now, a block of mass m slides at speed v on a smooth floor, and runs into a spring. At the greatest compression, how fast is the block moving? Zero, exactly! It stops for an instant, then gets pushed back. The floor is smooth, so friction takes nothing away. All the kinetic energy, half m v squared, is now stored in the spring as half k x squared. Okay? Cancel the halves: x squared equals m v squared over k. Take the square root, this tick-shaped sign, and x equals v times the square root of m over k. Remember the condition, though: a smooth floor. Clear?

3:24Vertical spring: hang it, or let it drop?

Now, a vertical spring. Hang a load of mass m on it, and let it settle. The spring's pull, k x naught, balances the weight, m g; g is the acceleration due to gravity. x naught, x with a small zero, is the extension at equilibrium. So x naught equals m g over k. Right? Now, attach the load to the spring at its natural length, and just let go. How far down does it go? x naught again, surely? No, no, careful! At x naught the forces balance, yes, but the load is still moving there. So it overshoots! It goes all the way down to two x naught, twice as far. We will check this with numbers in Example two. Okay? Now, the speed. Above x naught, the weight is bigger than the spring force, so the load speeds up. Below x naught, the spring wins, so it slows down. So the greatest speed is at x naught itself, the equilibrium position. Got it?

4:25The trap: lowering a load gently

Now the trap, and our puzzle. This time, you hold the load and lower it gently, until it hangs at rest at x naught. Gravitational energy lost: m g x naught. With me? But the spring stores half k x naught squared. And k x naught is m g, so that is half m g x naught. Only half! So where did the other half go? Into your hand, yes! Your hand pushes up while the load moves down, so the hand absorbs that energy. So there is no contradiction with energy conservation at all. Clear?

5:02Example 1: block lowered gently, where the other joule went

Now, remember our puzzle? Let's finish it. Example one: a two kilogram block, k g on screen, is lowered gently onto a spring of constant two hundred newtons per metre, written capital N, m to the power minus one. Find the extension, the energy stored, and the loss of gravitational potential energy. Take g equals ten. Pause the video, and try it first. Done? Okay, let's check. It hangs at rest, so x naught equals m g over k: two times ten, over two hundred. Zero point one metre, the small m, and that is ten centimetres. Okay? Energy stored: half times two hundred, times zero point one squared. One hundred times zero point zero one: one joule, capital J for joule. Right? Now the loss of gravitational energy. One joule as well, surely? No! m g x naught is two times ten times zero point one. Two joules, twice the energy stored! Exactly the trap we just saw. Clear? So the hand took the other one joule. Puzzle solved: the missing joule went into the hand! With me?

6:16Example 2: block released on a spring, max extension, speed

Now a slightly harder one, JEE level. Example two: the same two kilogram block is attached to the unstretched spring, and simply released. Find the greatest extension, and the greatest speed. Pause and try. Hint: at the lowest point, it is at rest for an instant. Done? Okay, let's check. Same spring, so k is two hundred newtons per metre, and g is ten. In Example one, the block hung at ten centimetres. So it stops at ten centimetres again, right? No, no, careful! Released suddenly, it overshoots. It starts at rest, and at the lowest point it is at rest again, so its kinetic energy does not change. So the gravitational energy lost, m g x, all goes into the spring as half k x squared. Okay? Cancel one x: x equals two m g over k. Two times two times ten, over two hundred: zero point two metre. Twenty centimetres, twice x naught, just as we said. Got it? Now the greatest speed. Where along the way is the block fastest? At x naught, yes! The equilibrium position, zero point one metre. There, the kinetic energy, half m v squared, equals the gravitational energy lost, m g x naught, minus the spring energy, half k x naught squared. Clear? From Example one, m g x naught is two joules, and the spring energy is one joule. So half times two times v squared equals two minus one, which is one. So v is one metre per second; m s to the minus one means metres per second. And see, this is the very joule the hand took in Example one. With no hand, it becomes kinetic energy! Beautiful, right?

8:04Quick revision: 1/2 kx^2, max compression, vertical spring

So, quick revision. Spring energy is half k x squared, or F squared over two k, or half F x, for stretch or compression. Block into a spring on a smooth floor: x equals v root m over k. Vertical spring: x naught is m g over k; released suddenly, it drops to two x naught, fastest at x naught. Lowered gently, only half of m g x naught is stored; the hand takes the rest. Now go and try the practice problems on this page. All the best!