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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.
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Example 1 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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Example 2 We
have a cable that weighs 2 lbs/ft attached to a bucket filled with coal that
weighs 800 lbs. The bucket is
initially at the bottom of a 500 ft mine shaft. Answer each of the following about this.
(a) Determine
the amount of work required to lift the bucket to the midpoint of
the shaft.
(b) Determine
the amount of work required to lift the bucket from the midpoint
of the shaft to the top of the shaft.
(c) Determine
the amount of work required to lift the bucket all the way up the
shaft.
Solution
Before answering either part we first need to determine
the force. In this case the force will
be the weight of the bucket and cable at any point in the shaft.
To determine a formula for this we will first need to set
a convention for x. For this problem we will set x to be the amount of cable that has
been pulled up. So at the bottom of
the shaft ,
at the midpoint of the shaft and at the top of the shaft . Also at any point in the shaft there is feet of cable still in the shaft.
The force then for any x
is then nothing more than the weight of the cable and bucket at that
point. This is,
We can now answer the questions.
(a) In this
case we want to know the work required to move the cable and bucket/coal from
to . The work required is,
(b) In this
case we want to move the cable and bucket/coal from to . The work required is,
(c) In this
case the work is,
Note that we could have instead just added the results
from the first two parts and we would have gotten the same answer to the
third part.
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Example 3 A
20 ft cable weighs 80 lbs and hangs from the ceiling of a building without
touching the floor. Determine the work
that must be done to lift the bottom end of the chain all the way up until it
touches the ceiling.
Solution
First we need to determine the weight per foot of the
cable. This is easy enough to get,
Next, let x be the distance from the ceiling to
any point on the cable. Using this
convention we can see that the portion of the cable in the range will actually be lifted. The portion of the cable in the range will not be lifted at all since once the
bottom of the cable has been lifted up to the ceiling the cable will be
doubled up and each portion will have a length of 10 ft. So, the upper 10 foot portion of the cable
will never be lifted while the lower 10 ft portion will be lifted.
Now, the very
bottom of the cable, ,
will be lifted 10 feet to get to the midpoint and then a further 10 feet to
get to the ceiling. A point 2 feet
from the bottom of the cable, will lift 8 feet to get to the midpoint and
then a further 8 feet until it reaches its final position (if it is 2 feet
from the bottom then its final position will be 2 feet from the ceiling). Continuing on in this fashion we can see
that for any point on the lower half of the cable, i.e. it will be lifted a total of .
As with the
previous example the force required to lift any point of the cable in this
range is simply the distance that point will be lifted times the weight/foot
of the cable. So, the force is then,
The work required is now,
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