Question regarding elastic potential energy and work

[gkangelexa]

Hi! This is probably something silly but here goes.

My question involves elastic potential energy and work…
So we know that a change in potential energy = Work done, as long as the forces are conservative...
delta U = Work done
Let’s say we have a spring…
Work done/by on a spring is, W= ½ kx^2
Also, the potential energy at a position on the spring is: U =½ kx^2

So if we have a spring with K = 360, the potential energy if you push it in 5 cm is: U = (1/2)(360)(.05)^2 = .45
If we then push it to 12 cm, U is now (1/2)(360)(.12)^2 = 2.59

So the difference in potential energy from 5 cm to 12 cm is 2.59-.45 which is 2.14.

But the  U = Work done
The work done to move from 5 cm to 12 cm should be equal to the difference in potential energy from position 5 cm to position 12 cm.
But when I calculate the work done to push the spring from 5 cm to 12 cm (a difference of 7 cm) it’s (1/2)(360)(.07)^2 = .882…
.882 does not equal 2.14…

When I try this method with gravitational potential energy (U = mgh; W = mgh) it works.
The work done to lift an object from a height of 5 m to a height of 12 cm is equal to the difference in potential energy from 5 to 12.

Why doesn’t it work with elastic potential energy just like it works with gravitational potential energy?

 

[tiny-tim]

hi gkangelexa!

(try using the X² icon just above the Reply box )

Work done/by on a spring is, W= ½ kx^2
…
But when I calculate the work done to push the spring from 5 cm to 12 cm (a difference of 7 cm) it’s (1/2)(360)(.07)^2 = .882…

no, .07 isn't x, it's ∆x

 

[gkangelexa]

So why does it work with gravitational potential energy?

The work done to lift an object from a height of 5 m to a height of 12 cm is equal to the difference in potential energy from 5 to 12.

 

[rock.freak667]

So why does it work with gravitational potential energy?

The work done to lift an object from a height of 5 m to a height of 12 cm is equal to the difference in potential energy from 5 to 12.

When you are calculating the change in something you are subtracting two values.

For elastic PE

ΔPE = PE₂-PE₁ = ½ kx₂² - ½ kx₁² = ½ k(x₂²-x₁²)

and you can see that 12²-5² ≠ 7

For gravitational PE:

ΔPE = PE₂-PE₁= mgh₂-mgh₁=mg(h₂-h₁)

and h₂-h₁ = 12-5 = 7.

For EPE you'd need to get the change in x² while for GPE you'd just need the change in h (or x).

 

[gkangelexa]

you guys are brilliant!