Work
In this section, we explore the relationship:
Work = Force · Distance
Remember that, in calculus, we often break the distances into very small
pieces-there is no surprise here-in calculating total work done, we will do the
same.
Example 1 Hooke’s Law states that the force (in Newtons) required to compress a spring is proportional to the distance it is being compressed. Find the
work done in compressing a spring 1/2 meter if the constant of proportionality is 8.
Solution.
The force function, F (x) = 8x. So, the total work done in
compressing the spring 1/2 meter is given by
Z 1/2
8xdx
W =
0
1/2
= 4x2 |0
= 1 joule
Theorem 1 In general, if force is a function of the position, x, that is,
F = F (x), then in moving from x = a to x = b, the work done is given by:
Z b
W =
F (x)dx
a
The Force Due to Gravity, Mass Versus Weight
One thing we should be aware of is that there is a difference between mass
and weight. The mass of a given object is the same everywhere...on earth, on
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the moon, on Jupiter-everywhere. The weight of that same object, however,
will change, depending on where the object exists. Weight takes into account
the force of gravity.
Note 1 A pound is a unit of weight (gravity is included here), whereas a
kilogram is a unit of mass.
When lifting an object vertically, gravity is acting on that object with
a downward force. We must ensure that gravity is accounted for in our
calculations.
If the weight, w, of an object is given, then the force due to gravity on
that object is w. If the mass, m, of an object is given, then the force due to
gravity on that object is mg = m(9.8m/s2 ).
Let’s look at some examples of calculating work done when an object is
being lifted.
Example 2 A 28-meter uniform chain, with mass 2 kilograms per meter is
dangling from the roof of a building. How much work is done in pulling the
chain up onto the roof ?
Solution. Note that we want to pull the chain up, vertically, to the top of
the building. Thus, we have the force of gravity opposing us. We were given
that the mass of the chain is 2kg/m. Notice that since we were given mass,
gravity is not accounted for. The downward Force per meter of chain, then
is given by
F = 9.8m/s2 (2kg/m)
= 19.6N
Then, the work done on moving a small slice (of length ∆y) of the chain y
meters will be
(19.6N )(∆y)(y meters)
Rearranging gives:
19.6y∆y
2
To calculate the total work done, we integrate
Z 28
19.6ydy
W =
0
= 9.8y 2 |28
0
= 7683.2joules
Example 3 Calculate the work done in pumping oil to the rim of the coneshaped tank having with diameter 25meters, and height 20meters, if the oil
has density 800kg/m3 and has vertical depth 10meters.
Solution. Since each horizontal slice of the oil must move the same distance
to reach the top of the tank, we will use horizontal slices. Then each slice is
a disk, with thickness ∆h.
Using similar triangles, we can find that the radius of such a disk will be
r = 58 h. Now, we calculate the force of gravity on a slice of the oil:
F = Density · Gravity · Volume
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= (800)(9.8)(π( h)2 ∆h
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Now, since each slice will need to travel a distance of (20 − h) meters to reach
the top of the tank, the work done on a slice will be
W = 3062.5πh2 (20 − h)∆h
Calculating the total work, we integrate
Z 10
W =
3062.5πh2 (20 − h)dh
0
Z
10
20h2 − h3 dh
0
20h3 h4 10
= 3062.5π
− |0
3
4
= 3062.5π
approx4 × 107 joules
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Example 4 The Great Pyramid of Egypt was built with stones having density of 200 pounds per cubic foot. The pyramid is 410 feet high, and has a
square base of 755ft. x 755ft. Calculate the work done in constructing the
Great Pyramid of Egypt.
Solution. Since we were given density in pounds per cubic foot, gravity is
already accounted for. Also, each slice of the pyramid is a square. Using
similar triangles, we can calculate the side length of a square at height h
(410 − h)ft. Then we can calculate the weight (that is,
as being s = 755
410
downward force) of a slice by multiplying by the density:
F = 200
2
755
(410 − h) ∆h
400
Now, we can calculate the work needed to lift a layer of rock:
2
755
W = 200
(410 − h) (h)∆h
400
To calculate total work, we integrate:
2
Z 410
755
(410 − h) (h)dh
200
W =
400
0
≈ 1.6 · 1012 foot-pounds
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