Calculus Compress Spring I have a question I cannot seem to get correct and am looking for some help.
Suppose a force of $40~\text{N}$ is required to compress a spring $3~\text{cm}$ from its equilibrium length. How much force is required to compress this an additional $5~\text{cm}$. (Recall $W = \int F dx$ and $F = kx$.)
What I did so far is
\begin{align*}
F & =kx\\
40 & =k(x)\\
W & = \int_{3}^{8} \frac{40x}{x} dx\\
W & = 320-120 = 200N
\end{align*}
I know this isn't the correct answer and I cannot figure out how to do it.
Thank you!
 A: The reason we are told that $40~\text{N}$ of force is required to compress the spring $3~\text{cm}$ is that this allows us to solve for $k$. 
We need to consider the units.  Since a Newton ($\text{N}$) is a unit of force and $F = ma$, where $m$ is the mass and $a$ is the acceleration,
$$\text{N} = \text{kg} \cdot \frac{\text{m}}{\text{s}^2}$$
where $\text{kg}$ stands for kilograms, $\text{m}$ stands for meters, and $\text{s}$ stands for seconds.  Hence, we need to express $3~\text{cm}$ as $0.03~\text{m}$.  
Using the equation $F = kx$ for the force applied to the spring to compress it $x~\text{m}$ gives 
$$k = \frac{F}{x} = \frac{40~\text{N}}{0.03~\text{m}} = \frac{4000}{3}~\frac{\text{N}}{\text{m}}$$
Hence, the force applied to the spring to compress it $x~\text{m}$ as a function of $x$ is 
$$F(x) = \left(\frac{4000}{3}~\frac{\text{N}}{\text{m}}\right)x$$
To determine the additional force needed to compress the spring from $3~\text{cm} = 0.03~\text{m}$ to $8~\text{cm} = 0.08~\text{m}$, you need to compute $F(0.08~\text{m}) - F(0.03~\text{m})$.  
