Project-based Calc II (1016-282-09)
1
Work
Work is defined as the quantity you get when you multiply the force in a given
direction by the distance it moves in that direction. For springs and other
systems where we can define a force as a function of position, F (x), we find that
work done moving an object from x = a to x = b is calculated by the expression
b
Z
F (x)dx
W =
(1)
a
This leads to an important concept that is often misunderstood: Work is defined as the difference between an integral evaluated at two points...you cannot
evaluate the “work at a point”, as this concept has no meaning whatsoever.
Interestingly enough, for lifting and pumping problems, this definition often
obscures things more than it helps. It says we calculate the force at some position, and integrate over distance. Instead, we can take the following approach:
Work out the differential amount of weight (force) at a given height, multiply
by the change in height we need to lift it, and then integrate. This is discussed
below.
2
Springs
We will work out the following two problems, trying the second in two different
ways, either of which is correct:
1. A force of 10 lbs. is required to hold a spring stretched 4 inches beyond
its natural length, How much work is done stretching it from its natural
length to six inches beyond its natural length? How much work is done
stretching it from two inches past it’s natural length to six inches past?
2. A spring has a natural length 20 cm. Compare the work W1 done in
stretching the spring from 20cm to 30cm with the work W2 done in stretching it from 30cm to 40cm. How are W1 and W2 related?
To do so, we will make use of Hooke’s law, which says that for any spring
not being really destroyed, the force required to hold it at some length x is
given by F = k(x − xn ), where xn is the “natural” length that results if no
force is applied, and k is some constant for the particular spring, with units of
[F orce/distance], typically N/m or lb/f t.
2.1
Spring problem 1
The first step in any problem is to get all the units done correctly from the
beginning. In English units, we work out everything in feet, so we convert 4
1
inches to 1/3 feet, 6in to 1/2 foot, and 2in to 1/6 foot. You will notice in the
problem that all distances are measured relative to the natural length, so we
can essentially choose that as our origin (i.e., choose coords such that xn = 0).
Now, we are given a force required to produce a given stretch, so we can use
Hooke’s law to determine k:
F = kx ⇒ k = F/x = (10 lb)/(1/3 ft) = 30 lbs/ft
(2)
Work is the integral of force over distance, and we now know the spring constant.
For part a.), we integrate from the natural length (x = 0) to six inches beyond
that (x = 1/2). We find
1/2
(1/2 ft)2 − 0
15
kx2 1/2
|0 = (30 lb/ft)
=
ft − lbs = 3.75 ft − lbs
2
2
4
0
(3)
For part b, the integrand is the same, but the bounds change. We need to
integrate from x = 1/6 ft to x = 1/2 ft, finding:
Z
kx dx =
W =
Z
1/2
W =
kx dx =
1/6
3
(1/2 ft)2 − (1/6 ft)2
kx2 1/2
|1/6 = (30 lb/ft)
= (30(2/9)/2) ft − lbs = (10/3) ft − lbs
2
2
(4)
Spring problem 2 - no change of variable
For spring problems where the natural length xn is not zero, we have two options.
Here, we demonstrate what happens if you just use the variables
you are given
R
originally. Note that since F (x) = k(x − xn ), we find W = k(x − xn )dx.
In the problem in question. the distances convert to xn = 20 cm = 0.2 m,
with 30cm= 0.3m, and 40cm= 0.4m. The work done to extend the spring from
20cm to 30cm (i.e., from x = 0.2m to x = 0.3m) is given by
Z
0.3
W1 =
2
0.3
k(x − xn )dx = k(x2 /2 − xn x)|0.3
0.2 = k(x − xn ) /2|0.2
(5)
0.2
Where k remains unknown, and the result you get depends on whether you
make an implicit u-substitution u = x − xn or not. Note that either of the two
rightmost expressions above are completely correct. The left one evaluates as
W1 = k([0.09 − 0.04]/2 − 0.2[0.3 − 0.2]) = k(0.025 − 0.02) = 0.005k. The right
one gives W1 = k(0.12 − 0)/2 = 0.005k, and is generally simpler.
We find the other term by similar means, using the general rule that work
for a spring is given by W = k(x − xn )2 /2|x=b
x=a , and find
Z
0.4
W2 =
2
2
k(x − xn )dx = k(x − xn )2 /2|0.4
0.3 = k(0.2 − 0.1 )/2 = 0.015k. (6)
0.3
Comparing, we find W2 = 3W1 .
2
3.1
Spring problem 2 - change of variables
We can also answer the second problem through a change of variables. Let
X be the spring’s extension away from its natural length of 20cm. The question
then states: A spring has a natural length X = 0. Compare the work W1 done
in stretching the spring from X = 0 to X = 10 cm = 0.1m with the work W2
done in stretching it from X = 10 cm = 0.1m to X = 20 cm = 0.2m. How are
W1 and W2 related?
R x=b
x=b
Now since F = kX in this case, W = x=a kXdX = kX 2 /2|x=a
. We find
Z
W1
0.1
2
kXdX = kX 2 /2|0.1
0 = k(0.1 )/2 = 0.005k
=
0
Z
W2
0.2
=
2
2
kXdX = kX 2 /2|0.2
0.1 = k(0.2 − 0.1 )/2 = 0.015k
0.1
W2
=
3W1
If you handle everything correctly, spring problems are easier this way!
4
Cable and lifting problems
For lifting problems, the key question is to ask how much mass/weight has to be
lifted/pumped just how far. Typically, these are easiest if you place the point
y = 0 at the top of your cable/tank, and work out how much mass lies at a
given height y. We’ll consider two examples:
4.1
Uniform density cable
A 500-foot cable has a linear density of ρl = 2lb/ft, and hangs vertically down
a shaft. How much work does it take to draw up the cable to the top of the
mineshaft?
Let y be the distance from the top of the shaft. How much work does it
take to take some amount of weight F (y) and bring it to the top of the shaft.
From our definitions, we would find W = yF (y). Unfortunately, the weight is
3
distributed over the length of the cable, so there is no single value to plug in for
the force. Instead, consider a tiny element of height dy. From the definitions,
this infinitesimal length of cable has a weight dF = ρl dy, i.e., the weight is equal
to the weight times the unit length multiplied by the infinitesimal segment of
length. This segment needs to be lifted ∆h = y units of height to reach the top
of the shaft, so the amount of work done is
Z
Z
Z
Z
W = F dy = ∆h(y)(dF ) = ∆h(y)ρl dy = yρl dy
(7)
This equation is correct for linear weight densities (lbs/ft). For a linear mass
density, we need to multiply by g for the result to be correct.
For the problem in question, we integrate the length of the cable from y = 0
to y = 500ft. Thus
Z 500
W =
ρl ydy = ρl y 2 /2|500
= 2(500)2 /2 = 250, 000 ft − lbs
(8)
0
0
You will notice that this is equivalent to lifting the entire weight of the cable,
1000 lbs, up a distance equal to half it’s length, 250ft, since that is the average
distance upward for any bit of the cable.
4.2
Non-uniform cable or other masses
Consider the slightly trickier case where the density of the cable varies linearly
from ρl = 1lb/ft at the top of the cable to ρl = 3lb/ft at the bottom. A quick
check will show that this cable too has a total weight of 1000lbs, but distributed
differently, with more weight near the bottom. As a function of y, it is easily
shown that ρl (y) = 1 + y/250. This time, the work to lift the entire cable is
Z
W =
h
Z
ρl ydy =
0
5
00(1+y/250)ydy = (y 2 /2+y 3 /750)|50 00 = 125, 000+400, 000/3 = 291, 666.66 lb − ft
0
(9)
If you put more mass at the bottom of the cable, it takes more work to lift it
since more of the mass has to travel further.
What if we attach the bottom of the cable to some heavy object? Here, it is
simple. We take the work required to lift the cable, and multiply by the weight
of the object times the height over which we lift it.
4
4.3
Pumping out a tank
In the case of pumping out a tank, we consider the same process. Let’s look
at a slightly harder case: A cylindrical tank with some radius R and height h
measured in meters, with a k meter high spout at the top. In effect, we need
to pump all of the water an extra meter to get it out. Let y be the height
of the top of the tank, ignoring the spout. What is the mass inside the tank
at some height y, measured downward from the top? Well, weight is given by
the expression F = mg, and we can find the mass through the density and
volume of a slice, m(y) = ρw V (y), so F (y) = gρw V (y). What is the volume
of an infinitesimal slice? Well, we multiply the cross-sectional area A(y) by the
infinitesimal height dy. We find F (y) = gρw A(y)dy. Now, how far does this
water need to be pumped upward? To get to the top of the tank, water travels
y meters, but the spout goes an extra k meters, so ∆h(y) = y + k. Now, the
cross-sectional area will be given by A(y) = πR2 . Thus, the work to pump the
tank out of the spout is given by
Z
W
=
Z
∆h(y)dF (y) =
h
Z
∆h(y)ρw gA(y)dy =
0
= πρw gR2
Z
h
ρw g(y + k)πR2 dy
0
h
(y + k)dy = πρw gR2 (y 2 /2 + ky)|h0 = πρw gR2 (h2 /2 + kh) = ρw g(πR2 h)[h/2 + k]
0
You’ll note the term in parentheses is the volume of the cylinder, and multiplied
by the density and g gives the weight of the water. We find the average height
traveled by this weight of water is equivalent to half the height of the cylinder
(this only holds for situations where the distribution is in some sesne symmetric,
not in general), plus an additional amount required to get the enitre weight of
water up the spout as well.
5
4.4
General rules for lifting/pumping problems
1. Define a height coordinate y
2. Identify the weight dF in an infinitesimal slice at height y via the formulae
• dF (y) = ρl dy or dF (y) = ρl gdy, for linear weight and mass densities,
respectively, or
• dF (y) = ρA(y)dy or dF (y) = ρgA(y)dy for volume-based weight and
mass densities, respectively
3. Figure how far the thing must be lifted/pumped from height y, and call
it ∆h(y).
R top
4. Integrate W = bottom ∆h(y)dF (y)
6