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: $$ W \approx \sum_{i=0}^{n-1} F(s_i) \Delta s_i \underset{n\to\inf}{\to} \int_{x_0}^{x_1}F(s) ds $$