Calculating Work on a Spring for Stretching from 4 to 7 Feet

[blessedcurse]

Homework Statement

The amount of WORK to stretch a spring 4 feet beyond its natural length of 2 feet is 10 ft-lbs. Find the work required to stretch the spring from 4 feet to 7 feet.

Homework Equations

W=\int^{b}_{a}Fdx=\int^{b}_{a}kxdx=[kx^{2}/2]^{b}_{a}

The Attempt at a Solution

10=[kx^{2}/2]^{4}_{0}
10=k(16)/2-k(0)/2
10=8k
k=4/5

W=[kx^{2}/2]^{b}_{a}
W=4/5[x^{2}/2]^{5}_{2}
W=4(25)/10-4(4)/10
W=42/5

W=8.4 ft*lb

 

[lanedance]

hi blessedcurse

this would probably go better in the physics forum...

But what is the force and energy stored in the spring at its natural length of 2ft?
F = kx applies when x is defined as the distance form equilibrium position...

 

[blessedcurse]

When the spring is at it's natural position of 2 feet, x=0, so F is 0 and W is 0...

Is that what you're asking?

 

[lanedance]

easier to update in new posts so i can keep track of what happens...

but yep - that looks better, first part is correct

there is some ambiguity in the 2nd part of the question whether it means:
~4ft-7ft total length (2 to 5 as you have done)
~4ft extension from natural poistion to 7ft (would give 4 to 7)

 

[blessedcurse]

I fixed my previous post before you had posted, I thought... sorry!

I don't know... I read it as stretching it from 2 to 5 feet beyond its natural length...