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### Compressing a Spring - Hooke's Law

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Date: 08/09/99 at 20:34:08
From: Frank Cozzone
Subject: Hooke's law

Assume that a force of 6 N is required to compress a spring from a
natural length of 4 m to a length of 3.5 m. Find the work required to
compress the spring from its natural length to a length of 2 m.

I have never done Hooke's law as compression. Could you please take me
through this step by step?
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Date: 08/12/99 at 17:11:59
From: Doctor Douglas
Subject: Re: Hooke's law

Hi Frank:

Here are the steps that I would use in solving this problem.

1) Write down the equation that expresses the physical principles that
are involved. In this problem, there seem to be two of them. One of
them deals with the properties of the spring, and you correctly
identify the principle as Hooke's law:

F = - k * x

Here F is the force that the spring exerts when the spring is either
stretched or compressed by a distance x (if x < 0, then that means
compression), and k is the so-called "spring constant." Springs like a
toy Slinky (TM) have small values of k, while stiff springs (like
shock absorbers on a car) have big values of k.

Now, the negative sign in Hooke's law means that the force that the
spring exerts is against the direction of the displacement. If you
attach one end of a spring to a fixed object and say that the other
end is at x = 0 when at its natural unstretched position, and then
stretch it in the +x direction by 2 m, then the force will be negative
(that is, it will be pulling back towards the -x direction). If you
compress the spring along the -x direction by 3 m, then it will push
back towards the +x direction as it tries to return to its natural
position at x = 0.  Does that make sense?

Okay, now that's one physical principle. The other seems to be the
definition of work. That is Integral[a,b]F dx, where I am using the
word Integral to mean the Integral sign, and the a and b are the
limits, viz: Integral[a,b] 3*x^2 dx = x^3 evaluated from a to b, or
b^3 - a^3.

2) With these principles in hand, we can now turn to recasting all
the words into equations:

"... a force of 6 N is required to compress a spring from a natural
length of 4 m to a length of 3.5 m"

So at x = (3.5m - 4.0m) = -0.5 m, the force exerted by the spring
pushes back with +6 N and counterbalances the applied 6 N force which
compressed it in the first place. In equations,

F = -k*x

+6 N = -k*(-0.5 m),

so

k = (6 N)/(0.5 m) = 12 N/m

This tells us exactly what the spring constant is. On to work:

"Find the work required to compress the spring from its natural length
to a length of 2 m."

Work required = Integral[a,b]Force-applied dx

natural length =  a = 4 m
ending length  =  b = 2 m

Force-applied  = -F = +k*x   (because F was defined previously
to be the spring force)

3) Now it's pretty straightforward to write down and compute the work
required as the integral.  I won't actually do it here, but if you
have any difficulty doing it, or with any of the steps that I've
outlined here, please feel free to write back!

track of the signs of various quantities. The easiest way to deal with
them is to set up a coordinate system in advance, as we did here: +x
means stretch along the +x direction, and hence the spring force of a
stretched spring is along -x, and so on.

Good luck!

- Doctor Douglas, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Physics/Chemistry

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