# Thread: Science Application of Integrals

1. ## Science Application of Integrals

A spring has a natural length of 10 in. An 800-lb force stretches the spring to 14 in.

A. Find the force constant.
work= force times displacement- though that applies here I do not know.

B. How much work is done in stretching the spring from 10 in to 12 in?
I have no idea.

C How far beyond its natural length will a 1600-lb force stretch the spring?
No idea.

All I understand is the boundaries are [10,14].

Thanks!

2. Originally Posted by Truthbetold
A spring has a natural length of 10 in. An 800-lb force stretches the spring to 14 in.

A. Find the force constant.
work= force times displacement- though that applies here I do not know.

B. How much work is done in stretching the spring from 10 in to 12 in?
I have no idea.

C How far beyond its natural length will a 1600-lb force stretch the spring?
No idea.

All I understand is the boundaries are [10,14].

Thanks!
Part (a):
This is an application of Hooke's Law $\displaystyle F=kx$. Solving for k, we get $\displaystyle k=\frac{F}{x}$. F is the force that stretches the spring (800 lbs) and x is the displacement of the spring from rest (4 in). Thus, $\displaystyle k=\frac{800lbs}{4in}=200\frac{lbs}{in}$.

Part (b):
Now that we have our k value, we can substitute into the equation for force:
$\displaystyle F=200x$.
If we integrate force with respect to x, we get work. The limits of integration now are from rest (x=0) to the displaced distance (x=2). The integral can be written as:
$\displaystyle W=200\int_{0}^{2}x\,dx=200(\frac{x^2}{2})_{0}^{2}= 400$.
It takes 400 Joules of work to displace the spring 2 inches.

Part (c)
Again, we would use Hooke's Law. Since we're dealing with the same spring, we would still have the same k value. Since $\displaystyle F=200x$ and $\displaystyle F=1600$, we see x (the displacement) is equal to $\displaystyle \frac{1600{lbs}}{200\frac{lbs}{in}}=8in$.
Thus, a 1600 lb force can displace the spring 8 in