# Total work compressing a conical spring

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 $$x=0.3$$.

$$dw = tF'$$ 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: $$W \approx \sum_{i=0}^{n-1} F(s_i) \Delta s_i \underset{n\to\inf}{\to} \int_{x_0}^{x_1}F(s) ds$$
• (Oh, I used $s$ instead of your $t$, since $s$ is usually distance). Feb 22, 2021 at 10:05
• 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 Feb 22, 2021 at 19:20
• 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 $W\ge 45\cdot0.3=13.5$. Feb 22, 2021 at 20:36
• Incidentally, since the force changes linearly, we can actually just use the average of the two boundary values: $W=(150+45)/2\cdot0.3$. Feb 22, 2021 at 20:40