Springs and Work

[eugenebrooks]

[quote=sanddrag]I assume you meant 42 instead of 40.
[\QUOTE]

I don't think so. The stretch from 30 to 42 centimeters
requires a work of 2 joules. Given the formula for
work, W = (a/2) stretch^2 you calculate a, or if you
are hung up on variable names, k.
The result, using the data you have been given, is 4/144.
Don't get hung up on the units, it is okay to use joules
per cm^2 as long as you are consistent...

You are then asked for the work to stretch from 35 to 40
cm, and you can get this by subtracting the work required
to stretch from 30 to 35, W1, from the work required to stretch
from 30 to 40, W2.

W2 = (4/144) * (1/2) * (40-30)^2
W1 = (4/144) * (1/2) * (35-30)^2

W2-W1 = (2/144) * (100 - 25) = 150/144 = 25/24 joules

This is how a physicist views the world,
for how an engineer views it refer to Paul's
correctly done calculation. Either view
can be used to solve the problem...

[Denman]

from my double maths

EPE (elastic potential energy / work done from natural length) = lambda x^2 /2l
i've attached a spreadsheet
first, work out value of the spring force thing (grr can't remember proper name)
even better you can work out lambda/2l
from the working you can see you get this to be 2/144
then use this value to work out the work done in extending the spring from the point with no extention, to x=5 and x=10, and work out the difference

i get 25/24, or1.041667
Hope this helps