Work
This is the final application of integral that we�ll be
looking at in this course. In this section we will be looking at the amount of
work that is done by a force in moving an object.
In a first course in Physics you typically look at the work
that a constant force, F, does when moving an object over a distance of
d. In these cases the work is,
However, most forces are not constant and will depend upon
where exactly the force is acting. So, let�s suppose that the force at any x
is given by F(x). Then the work done by the force in moving an object
from
 to
 is
given by,
To see a justification of this formula see the
Proof of Various Integral Properties section of the Extras chapter.
Notice that if the force constant we get the correct
formula for a constant force.
where b-a is simply the distance moved, or d.
So, let�s take a look at a couple of examples of
non-constant forces.
Example
A spring has a natural length of 20 cm. A 40 N force is required to
stretch (and hold the spring) to a length of 30 cm. How much work is done in
stretching the spring from 35 cm to 38 cm?
Solution
This example will require Hooke�s Law to determine the
force. Hooke�s Law tells us that the force required to stretch a spring a
distance of x meters from its natural length is,
where
 is
called the spring constant.
The first thing that we need to do is determine the spring
constant for this spring. We can do that using the initial information. A
force of 40 N is required to stretch the spring 30cm-20cm = 10cm = 0.10m from
its natural length. Using Hooke�s Law we have,
So, according to Hooke�s Law the force required to hold
this spring x meters from its natural length is,
We want to know the work required to stretch the spring
from 35cm to 38cm. First we need to convert these into distances from the
natural length in meters. Doing that gives us x�s of 0.15m and 0.18m.
The work is then,
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