Thread: 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!

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 . Solving for k, we get . F is the force that stretches the spring (800 lbs) and x is the displacement of the spring from rest (4 in). Thus, .


Part (b):
Now that we have our k value, we can substitute into the equation for force:
.
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:
.
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 and , we see x (the displacement) is equal to .
Thus, a 1600 lb force can displace the spring 8 in

Hope that answers your question!!!!

You have the correct answer, but be warned. The unit for work here is NOT Joules! It is lb in. (The typical unit here would be lb ft.)

-Dan

Originally Posted by topsquark

You have the correct answer, but be warned. The unit for work here is NOT Joules! It is lb in. (The typical unit here would be lb ft.)

-Dan

Thanks for catching that!