How Is the Spring Constant Calculated from Work Done?

[bob tran]

Homework Statement

It requires 49 J of work to stretch an ideal very light spring from a length of 1.4 m to a length of 2.9 m. What is the value of the spring constant of this spring?

Homework Equations

u=0.5kΔx²

The Attempt at a Solution

[tex]
u=\frac{1}{2}k\Delta x^2\\
k=\frac{2u}{\Delta x^2}\\
k=\frac{2*49}{(2.9-1.4)^2}\\
k=44 \ \texttt{N}/\texttt{m}
[/tex]
Apparently, the answer is 15 N/m.

EDIT:
It appears they have done this instead:
[tex]
u=\frac{1}{2}kx_f^2 - \frac{1}{2}kx_0^2\\
u=\frac{1}{2}k(x_f^2-x_0^2)\\
k=\frac{2u}{(x_f^2-x_0^2)}\\
k=\frac{2*49}{(2.9^2-1.4^2)}\\
k=15 \ \texttt{N}/\texttt{m}
[/tex]
I thought Δx would be squared after the subtraction?

 

[SteamKing]

Homework Statement

It requires 49 J of work to stretch an ideal very light spring from a length of 1.4 m to a length of 2.9 m. What is the value of the spring constant of this spring?

Homework Equations

u=0.5kΔx²

The Attempt at a Solution

[tex]
u=\frac{1}{2}k\Delta x^2\\
k=\frac{2u}{\Delta x^2}\\
k=\frac{2*49}{(2.9-1.4)^2}\\
k=44 \ \texttt{N}/\texttt{m}
[/tex]
Apparently, the answer is 15 N/m.

EDIT:
It appears they have done this instead:
[tex]
u=\frac{1}{2}kx_f^2 - \frac{1}{2}kx_0^2\\
u=\frac{1}{2}k(x_f^2-x_0^2)\\
k=\frac{2u}{(x_f^2-x_0^2)}\\
k=\frac{2*49}{(2.9^2-1.4^2)}\\
k=15 \ \texttt{N}/\texttt{m}
[/tex]
I thought Δx would be squared after the subtraction?

You aren't given the unstretched length of the spring, only that it is stretched from a length of 1.4 m to 2.9 m.

The energy it takes to stretch a spring, namely E = (1/2)k x², is applicable only for a spring which is stretched a distance x from its unstretched length.

http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html

 

[Bengalfan985]

The equation you used is only for when the spring is at rest, Δx is calculated from the equilibrium point. You're work would be correct if they said the spring was stretched 1.5 meters from it's equilibrium point.

Think about a spring that is as rest compared to a similar one that is stretched quiet far. The force exerted by an ideal spring increases as the spring gets further away from it's equilibrium point. If both are stretched 1.5 meters, the one that's already stretched quiet far would be harder to stretch than the one that was at rest. The equation you used would give the same answer for both springs.

The equation under "EDIT" is the one you might want to always think of, and just assign zero to the second "x" if the spring is at rest.

 

[Chestermiller]

Bob Tran: As SteamKing points out, their answer is not correct either, unless the initial length of the spring is zero (which is silly). I think your original approach is a better interpretation of the problem statement than the given answer (although the problem statement is clearly underspecified).

Chet

 

[gneill]

One can approach the problem in terms of the definition of work done:
$$ W = \int_{x_o}^{x_f} f(x)~dx $$
$$ W = \int_{x_o}^{x_f} k~x~dx $$
and arrive at the formula and result that the book gives.

 

[SammyS]

@bob tran ,
What book is this problem from?

 

[Chestermiller]

One can approach the problem in terms of the definition of work done:
$$ W = \int_{x_o}^{x_f} f(x)~dx $$
$$ W = \int_{x_o}^{x_f} k~x~dx $$
and arrive at the formula and result that the book gives.

This is correct only if the unextended length is zero, right?

Chet

 

[gneill]

This is correct only if the unextended length is zero, right?

Chet

I believe it should apply over any segment of the stretch.

 

[SammyS]

Chet is correct.

 

[gneill]

Chet is correct.

You're right. My bad

 

[bob tran]

@bob tran ,
What book is this problem from?

Not sure. It was from a practice worksheet.