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A conical spring's stiffness varies linearly with displacement from rest. Its stiffness when uncompressed is 45 N/m, and 150 N/m when fully compressed. The uncompressed spring stretches 30 cm further than the compressed strength. Find the total work compressing the spring.


The first thing I did was outline my total work formula. The total work from $t=0$ to $t=x$ (in meters), $W(0, x)$, is $$\int_0^xr_w(t)dt$$ where $r_w(t)$ is the rate of change of work. I know that work is an amount of force (which varies) applied over some distance, $w=Fd$, so then rate of change of work is $dw=dF'=tF'$. The rate of change of force is given by $$\biggr(\frac{150-45}{0.3}\biggr)t+45=350t+45$$

Then $$r_w(t)=t(350t+45)$$ Which follows that \begin{align*}W(0,x)&=\int_0^x(350t^2+45t)dt \\ W(0, 0.3)&= 5.175\end{align*} I believe this is in N/m. Is this correct? It sounds way too low for me given the rates provided at $x=0$ and $x=0.3$.

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$dw = tF'$ is wrong. It should be $dW = F(s)ds$, so $$ W = \int_{x_0}^{x_1}F(s) ds $$ Note that $F(s)ds$ is force times distance, so it generalizes $W=Fd$.

Also the unit isn't $N/m$, it's $Nm=J$, once again because we multiply $F$ and $ds$ (in a manner of speaking).

BONUS: To find the correct integral yourself, you can approximate by assuming constant $F$ over small distances. Divide the interval $[x_0,x_1]$ into $n$ small intervals $[s_i,s_{i+1}]$ for $i=0\ldots n$. Then: $$ W \approx \sum_{i=0}^{n-1} F(s_i) \Delta s_i \underset{n\to\inf}{\to} \int_{x_0}^{x_1}F(s) ds $$

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  • $\begingroup$ (Oh, I used $s$ instead of your $t$, since $s$ is usually distance). $\endgroup$
    – Milten
    Feb 22, 2021 at 10:05
  • $\begingroup$ Interestingly, my homework responded that the integral I provided is correct. I just can't make sense of the solution here because I'm not exactly sure how to define $F(s)$. I have a linear equation describing force, but the multiple choice answers require another $s$ be multiplied like I did in my post $\endgroup$
    – Lex_i
    Feb 22, 2021 at 19:20
  • $\begingroup$ Your linear expression for the force is correct, so you should just integrate that (without multiplying by $t$). You do write that the expression is the rate of change of the force — that’s wrong, it’s just the force. Also you can sanity check a final answer by using the lower and upper bounds of the force. E.g. a lower bound is $W\ge 45\cdot0.3=13.5$. $\endgroup$
    – Milten
    Feb 22, 2021 at 20:36
  • $\begingroup$ Incidentally, since the force changes linearly, we can actually just use the average of the two boundary values: $W=(150+45)/2\cdot0.3$. $\endgroup$
    – Milten
    Feb 22, 2021 at 20:40

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