Energy stored in a stretched wire We know

Energy stored in a stretched wire We know that generally: Considering the force extension graph shown opposite, the extension, Δx, is produced by an average force of F/2. So the work done in stretching the wire is … F Force ΔW = F Δx Extension Δx ΔW = ½ F Δx 1 © TPS 2008

Energy stored in a stretched wire This is the elastic energy stored in the stretched wire. We could, more generally, write: F Force ΔW = ½ F Δx Eel = ½ Fx Where: Extension Δx • Eel is the elastic strain energy stored in the wire • F is the force applied to stretch the wire • x is the extension in the wire 2

Energy stored in a stretched wire Eel = ½ Fx We specifically mentioned a wire when looking at this formula but it holds true for anything that can be either compressed or stretched and obeys Hooke’s law*. *This means that whatever is being stretched has not gone beyond its limit of proportionality. 3

Energy stored in a stretched wire Example A spring with a spring constant (k) of 100 Nm-1 has a weight of 14 N gently lowered onto it. How much energy is stored in the spring when it has reached an equilibrium position? Solution We wish to use the formula Eel = ½ Fx but we do not know the extension, x, caused by the 14 N weight. However, we can calculate this from the spring constant. F = kx So x = F/k And x = 14 / 100 = 0. 14 m 4

Energy stored in a stretched wire Example A spring with a spring constant (k) of 100 Nm-1 has a weight of 14 N gently lowered onto it. How much energy is stored in the spring when it has reached an equilibrium position? Solution So Eel = ½ x 14 x 0. 14 1 J of energy is stored in the stretched spring. I. e. Eel = 1. 0 J 0. 14 m Now we can use the formula Eel = ½ Fx. The spring has 1 J of energy stored within it, due to its extension. 14 N How much gravitational potential energy has the weight on the end of the spring lost? 5

Energy stored in a stretched wire As the weight on the spring is lowered to the equilibrium position, it will lose gravitational potential energy, ΔEgrav, given by the equation: ΔEgrav = mgΔh (where m is the mass, Δh is the change in height and g is the gravitational field strength. ) 1 J of energy Note that mg is the weight of the body on is stored in the end of the spring, 14 N. the stretched spring. So we get ΔEgrav = 14 x 0. 14 = 2 J 0. 14 m Half of this, 1 J, is stored in the spring; where does the rest go? 14 N 6

Energy stored in a stretched wire Work is done on your hand as you gently lower the weight to the equilibrium position. If you find this hard to envisage, imagine whet would happen if your hand wasn’t there. Suppose you just attached the 14 N weight to the spring and allowed it to fall. The spring would extend beyond 0. 14 m and then bounce back up. When the extension was 0. 14 m, the weight would still be moving – it would have kinetic energy along with the energy stored in the spring. 1 J of energy is stored in the stretched spring. 0. 14 m 14 N The 2 J change in gravitational potential energy would be converted 7 to the 1 J of kinetic energy and 1 J elastic potential energy.

Energy stored in a stretched wire Additionally, we can see that it relates directly to the force extension graph. Force Eel = ½ Fx is a convenient formula to use. F Extension x You will remember that the formula for the area of a triangle is Area = ½ Base x Height Now look at the force extension graph. The area of the triangle marked in yellow is … ½ Fx which is … … the energy stored in the stretched spring. The energy stored in a stretched spring is equal to the area … under the curve of a force extension graph. 8

The energy stored in a stretched spring is equal to the area under the curve of a force extension graph. We shall now look at three examples to illustrate three points. Force Energy stored in a stretched wire F Extension x This holds true for any force extension curve (even if it is not a straight line!). There are three important things to remember: 1. The area is calculated by the units of force multiplied by extension 2. The extension is the whole extension, not just that due to an additional load 9 3. The area under a curve can be calculated by “counting the squares”

Energy stored in a stretched wire The area is calculated using the units of force and extension Example 1 – assume the spring obeys Hooke’s Law Calculate the energy stored in a spring which has been loaded with a 5 N weight and stretched to 25 cm. Eel = area under the curve Area = ½ x 25 cm x 5 N Area = 62. 5 Ncm Area = 0. 625 Nm 4 Force / N Area = ½ base x height 2 5 10 15 Extension / cm 20 25 Eel = 0. 625 J 10

Energy stored in a stretched wire The area is calculated using the units of force and extension Example 2 – assume the spring obeys Hooke’s Law Calculate the extra energy stored in the spring as an additional 2 N weight is added, extending the spring to 35 cm. For 5 N load: Eel = area under the curve For 7 N load: Area = ½ base x height Area = ½ x 25 cm x 5 N Area = ½ x 35 cm x 7 N Area = 62. 5 Ncm Area = 122. 5 Ncm Area = 0. 625 Nm Area = 1. 225 Nm Eel = 0. 625 J Eel = 1. 225 J So extra energy stored = 1. 225 -0. 625 = 0. 6 J 11

Energy stored in a stretched wire The area is calculated using the units of force and extension Example 2 Total area is equivalent to 1. 225 J 6 Force / N 4 0. 6 J 2 0. 625 J 5 10 15 Extension / cm 20 25 30 35 To find the extra energy stored, we have first found the total energy stored and then subtracted the energy stored when it is extended by 25 cm. 12

Energy stored in a stretched wire The area is calculated using the units of force and extension Example 2 We could have calculated the area shaded in red instead. ½ x 10 cm x 2 N 6 Force / N 4 2 10 cm x 5 N 0. 625 J 5 10 15 Extension / cm 20 25 30 35 Area = (10 cm x 5 N) + (½ x 10 cm x 2 N) = 50 Ncm + 10 Ncm = 60 Ncm So energy stored is 0. 6 J 13

Energy stored in a stretched wire The area is calculated using the units of force and extension Example 3 –this sample does not obey Hooke’s Law Calculate the energy stored in the sample when it has been loaded to extend by 25 cm. Eel = area under the curve 11 4 Force / N Divide the area up into rectangles 10 of 5 cm x 1 N i. e. of area 5 Ncm. 6 0. 5 2 9 5 Now match up the incomplete 1 2 12 rectangles. 5 This leaves an estimated ½ of a 10 15 Extension / cm rectangle. So in total there are 12. 5 rectangles each representing 5 Ncm. This 12. 5 x 5 Ncm = 62. 5 Ncm or 0. 625 Nm So the energy stored is 0. 625 J. 8 7 4 3 20 25 14

Energy stored in a stretched wire Not all stretched wires return to their original length. We shall now consider what happens to the energy stored in a stretched wire, as it is stretched beyond its elastic limit (the point beyond which it will not return to its original length if unloaded). Force The energy stored in the wire is calculated from the total area under the curve. Force Extension When the wire is unloaded, it does not return to its original length, as its structure has been changed. The energy released on unloading will be given by the area under the unloading curve. This is marked in yellow in the diagram. So, where has the energy gone that is represented by the area under the rest of the curve? 15

Force Energy stored in a stretched wire Extension Remember, the total area under the loading curve represented the energy “stored” during stretching the wire. However, the yellow area shows the stored energy released on unloading. The remaining energy, represented by the striped area under the curve has been retained in the wire. The energy has been used to deform the structure of the wire, allowing for its permanent extension. This will occur whenever a sample is stretched beyond its elastic limit. 16

Energy stored in a stretched wire Force You may remember that when you stretch a rubber band, it exhibits hysteresis. The unloading curve is not the same as the loading curve, even though it returns to the original length. Extension 17

Energy stored in a stretched wire Force When loading the rubber band, the “energy stored” in the rubber band is represented by the area under the curve. Extension 18

Energy stored in a stretched wire Force The energy released as the rubber is unloaded will be given by the area under the unloading curve. This is shown in yellow in this diagram. Extension 19

Energy stored in a stretched wire Force What has happened to the energy that is represented by the remaining area, shaded in pale blue, between the two curves? Extension 20

Energy stored in a stretched wire Force Remember, there is no overall, permanent extension to the rubber band so energy cannot have been stored up in permanently deforming the structure of the rubber. Extension 21

Energy stored in a stretched wire Force However, the energy has been dissipated as heat. Try it for yourself. Repeatedly stretch and contract a rubber band it will get warm. This heat energy is represented by the area between the curves. Extension 22

Force Energy stored in a stretched wire Extension So, the area under the curve in this graph, for a rubber band, represents more than the energy stored in the band. It also includes the heat energy dissipated as the band is stretched. This energy cannot be regained by unloading the rubber band. 23

Energy stored in a stretched wire Force And the area under this curve for a stretched wire, represents the energy stored in the wire due to both its deformation as well as energy that can be returned by unloading the wire. Extension 24

Energy stored in a stretched wire We have looked at what happens when we stretch a wire and an elastic band. Clearly, there are many other substances that can be investigated. The equation that we have used will work for all the other substances, both for compressive and tensile stresses. 25

Energy stored in a stretched wire This includes gases. The behaviour of gases is simple to investigate by using a syringe of gas but take care, especially if the syringe is made of glass. You must, of course, block the end of the syringe effectively. 26

Energy stored in a stretched wire Liquids are often quoted as being incompressible. This is a question of degree and obviously, with sufficient pressure, more or less anything can be compressed. However, water, for example, could not be compressed to a measurable extent in the school laboratory. 27

Energy stored in a stretched wire There are many solids which would be interesting to investigate, such as brick and concrete but again, the changes in length are so small as to be difficult to measure in the school laboratory. 28

Energy stored in a stretched wire Applying tensile stresses to water in the laboratory is also likely to cause problems because: • Dissolved gasses are likely to evaporate, making measurements inappropriate • Any change in volume would be so small as to be immeasurable with the available equipment 29