Question

Can anyone please help. Two springs, with force constants k1 and k2, are connected in series. How much work is required to stretch this system a distance x from the equilibrium position?
Express your answer in terms of the given quantities.
Thank You

Answer 1

Here is the figure:

Answer 2

$$
F_1=k_1x\\
F_2=k_2x\\
\text{Total Force }=F_1+F_2\\
\text{Total Work (force times distance) }=\left(F_1+F_2\right)x=\left (k_1+k_2\right )x^2\text{ Joules}\\
$$
Does this make sense?

Answer 3

I tried that and I still got it wrong. I also tried many other ways, but it keeps marking it wrong. Thanks for the help anyway. There is probably something wrong with the homework website.

Answer 4

Actually, the springs would have to be in PARALLEL for you to just add their individual spring constants. For springs in series, the value of the new k you get is given by this equation:

\[\frac{1}{k_{new}}=\frac{1}{k_{1}}+\frac{1}{k_{2}}\]
Once you find this new k (the equivalent k), use that in the W=kx^2 equation!

Answer 5

I just found that on the wikipedia page it shows the derivations for the equations for parallel and series situations. Have a look through if you want to see why the equations are the way they are!

http://en.wikipedia.org/wiki/Series_and_parallel_springs

Answer 6

Yes, that makes sense!
Total displacement of spring 1 plus spring 2 is
$$
x=x_1+x_2
$$
The force on each spring is
$$
F=k_1x_1\\
\implies x_1={F\over k_1}\\
F=k_2x_2\\
\implies x_2={F\over k_2}\\
F=k_{eff}x\\
{\cancel{F}\over k_{eff}}=x_1+x_2={\cancel{F}\over k_1}+{\cancel{F}\over k_2}\\
k_{eff}={{k_1k_2\over k_1+k_2}}\\

$$
Therefore, total work is
$$
Fx=k_{eff}x^2={{k_1k_2\over k_1+k_2}}x^2\\
$$

Answer 7

W = F • x but when F = f(x) you need to integrate

For a spring, work = 1/2 k x^2 = energy stored on spring.

Answer 8

Yes @IrishBoy123 is absolutely right! Thanks for that! (was almost 2 am when I posted the answer...)

@autogenius

Answer 9

Thanks for the help but how did you get the numerator k1k2?

Answer 10

\[\frac{ 1 }{ a } + \frac{ 1 }{ b } = \frac {a + b}{ab}\]

Answer 11

He just rearranged the equation I wrote to solve for the equivalent k, k(new) or k(e):

\[{1 \over k_{e}} = {1 \over k_1} + {1 \over k_2}\]
Multiply equation by k(e)k(1)k(2):
\[{k_e k_1 k_2 \over k_e} = {k_e k_1 k_2 \over k_1}+{k_e k_1 k_2 \over k_2}\]
\[{k_1 k_2} = {k_e k_2 }+{k_e k_1}\]
\[k_1 k_2=k_e(k_2+k_1)\]
\[k_e = {k_1 k_2 \over k_1 + k_2}\]

Answer 12

@IrishBoy123 - thanks for reminding us F=f(x) !
Since F is varying as a function of time we need to to account for the fact that F is not constant when taking the \(\bf{area}\) of the F vs. x curve, which represents \(\it{work}\):

Or using calculus,
$$
\large{
\int_0^{x_c}f(x)dx=\int_0^{x_c}k_{eff}xdx={1\over2}k_{eff}x^2
}
$$