Note-A-Rific: Elastic Potential Energy


Any object than can be deformed (have its shaped changed) and then return to its original shape can store elastic potential energy.

·        We’re still talking about potential energy, since it is stored energy until the object is allowed to bounce back.

·        “Elastic” does not refer to just things like elastic bands…other materials that would be referred to as elastic would be

o       pole vaulter’s pole

o       springs

o       cheese (no, I’m just kidding. Just wanted to see if you’re paying attention)

 

You learned in Physics 20 that Hooke’s Law is…

F = kx

F = force (N)

k = spring constant for that object (N/m)

x = amount of expansion or compression (m)

 

We can use this formula to figure out a formula for the energy stored in the spring.

·        Remember that W = F d

·        We might be tempted to just shove the formula for Hooke’s Law into this formula to get W  = kxd = kx2, but this is wrong!

·        You have to take into account that the force is not constant as the object returns to its original shape… it’s at a maximum when it is deformed the most, and is zero when the obect is not deformed.

·        Let’s graph Force vs Distance of Expansion for a spring that was stretched and we are now letting go of it…

 

 

 

 

 

 

 

 

 

 

 

 

 

 


But this is really just a Force vs Displacement Graph like the ones we just looked at a couple of sections back! To figure out the energy of the spring we can just figure out the work it does by looking at the area under the graph.

 

Area = ½ bh

            = ½ F x

            = ½ (kx) x

Area = ½ kx2 = W

 

So the work done by the spring (and then energy it stored) can be calculated using…

 

            Ee = ½ kx2

                                    Ee = eleastic potential energy (J)

                                    k = spring constant (N/m)

                                    x = amount of expansion or compression [deformation] (m)

 

 

 

 

Example: How much energy does a spring with a k constant of 15 N/m store if it is stretched by 1.6m?

Ee = ½ kx2

            = ½ (15N/m) (1.6 m)2

Ee = 19 J