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Compressing a Spring - Hooke's Law


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?


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!

One of the most confusing things about this type of problem is keeping 
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/   
    
Associated Topics:
High School Physics/Chemistry

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