Question

Find the natural length of a heavy metal spring, given that the work done in stretching it from a length of 2ft to 3ft is one half the work done in stretching it from 3ft to 4ft.

Answer 1

@freckles would you mind helping me out with this problem?

Answer 2

the energy stored in the spring is given by \(E = \frac{1}{2} k (x - x_o)^2\)

where \(x_0\) is it's natural length

the energy stored is the work done. from here it's just a maths problem.

so "the work done in stretching it from a length of 2ft to 3ft " is it's concomitant increase in energy which is \(\frac{1}{2} k (3 - x_o)^2 - \frac{1}{2} k (2 - x_o)^2\)

pattern match for stretching it from a length of 3ft to 4ft

k's cancel, so looks like it's just algebra from there, ie solve for \(x_o\) :p

Answer 3

Do you know is there is a way to solve this with integration?

Answer 4

So according ti your way, would I set up the problem like this?

\[\frac{ 1 }{ 2 }k(3-x _{0})^{2}-\frac{ 1 }{ 2 }k(2-x _{0})^{2} = \frac{ 1 }{ 2 }[\frac{ 1 }{ 2 }k(4-x _{0})^{2}-\frac{ 1 }{ 2 }k(3-x _{0})^{2}]\]

Answer 5

@IrishBoy123

Answer 6

does that not work for you?

i'd cancel the k's

leaving one equation, 1 unknown, ie \(x_o\)

bish bosh!

Answer 7

I'm having difficulty understanding what you're trying to say. So would I have to set the equation that you gave me to zero and then solve for \[x _{0}\]?

Answer 8

@IrishBoy123

Answer 9

\[\frac{ 1 }{ 2 }k(3-x _{0})^{2}-\frac{ 1 }{ 2 }k(2-x _{0})^{2} = \frac{ 1 }{ 2 }[\frac{ 1 }{ 2 }k(4-x _{0})^{2}-\frac{ 1 }{ 2 }k(3-x _{0})^{2}]\]

\[(3-x _{0})^{2}-(2-x _{0})^{2} = \frac{ 1 }{ 2 }[(4-x _{0})^{2}- (3-x _{0})^{2}]\]

that's a quadratic

Answer 10

Ok, thank you!