Download Contrary to everyday usage, the term work has a very specific

Work
Contrary to everyday usage, the term work has a very specific meaning in physics. In physics,
work is related to the transfer of energy by forces. There are two different ways to conceive
of work, and the usefulness of each approach depends on the situation. According to the
work-kinetic energy theorem, the net work done on a object is equal to the object’s change
in kinetic energy (energy associated with motion).
Wnet = ∆KE = KEf − KEi
Similarly, we can consider the work done only by nonconservative forces in a system, and we
find that the total work done on an object by nonconservative forces is equal to the object’s
change in total energy (kinetic plus potential energy).
Let us apply these ideas in a few situations in order to get a better understanding of them.
For example, we can think of throwing a ball. When an individual throws a ball he or she
exerts a (nonconservative) force on the ball, which sends the ball into motion through the
air. Energy is transfered in terms of kinetic energy, related to the motion of the ball through
air. Since energy is transfered into the ball by the force of throwing the ball, the individual
throwing the ball is doing work on the ball.
If we restrict ourselves to one-dimensional motion, we find the work done by a constant force
acting on an object to be given by
W = Fd
where F is the force and d is the displacement of the object (the amount it is moved by the
force). It follows that the units of work are Newton meters (Nm), which are better known
as Joules (J), the unit of energy. Since the units of work are energy, we can see that the
amount of work done by a force on an object is simply the amount of energy transfered by
the force into the object.
It is worth noting that simply because a force acts on a given object, it does not necessarily
do work on the given object. Think about a situation where a desk is so heavy, that no
matter how hard you try to push it, it does not move. In this situation you are clearly
exerting a force on the desk, yet you are doing no work on the desk because the desk has
zero displacement. Thus, in pushing on the desk this way no energy is transfered into the
desk.
For a third example, we can think about lifting a box vertically, to say a height h. In order
to lift the box we will need to exert a force opposite to that of gravity which is pushing the
box downward. If we lift the box slowly, using a constant force about equal in magnitude to
the weight of the box (the force due to gravity), the work done by lifting the box will be
Wl = F d = mgh
However, at the same time gravity is acting with a force of −mg, so
Wg = Fg d = −mgh
and so no work is done overall. This means that there is no change the object’s kinetic
energy, which makes sense, because after the object has been lifted a height h it is at rest.
However, if we consider only the nonconservative forces acting in this situation, we can gain
some additional information. Gravity is a conservative force, so we neglect the work done
by it. Then, we see that the net work done by nonconservative forces is mgh, which corresponds to a change in total energy, but not kinetic energy. Thus, this change in energy must
be related to the potential energy of the box. After being lifted, the box has gravitational
potential energy, related to its increased height. In this situation work is done by the person
on the box, which increases the energy of the box, but in doing work the energy of the
person is decreased. If the box is dropped, then it will begin to accelerate and have energy
corresponding with motion. When it finally hits the ground energy will be transfered into
heat and sound.
In a more general situation, we can consider the amount of work done on an object by a force
that is not constant. Think of stretching out a spring. The further the spring is stretched,
the more force (and thus work) required to move it an additional distance. In order to find
the total work required to stretch the spring a given distance, we can think of subdividing the
distance the spring is stretched into very small subintervals. As these subintervals become
small enough, or of length dx, the force required to stretch the spring the distance dx will
be relatively constant, because the resistance of the spring will not change over such a small
distance. To find the total work required to stretch the spring a distance b − a, we simply
sum the work required to stretch the spring each infinitessimal distance. Thus, for a variable
force, we find that the work done in moving an object from a to b is given by
Z b
W =
F (x)dx
a
Hooke’s law says when a spring is stretched or compressed a distance x from its natural
(uncompressed) state, it exerts a force of
Fs = −kx
where k is a spring constant, related to the specific design of the spring. This force is a a
restoring force, which tends to push the spring to equilibrium. Thus, in order to stretch or
compress a spring a distance x, one must exert a force of
Fa = kx
so that if this force is maintained it will balance with the restoring force, and the spring will
remain at a position x.
Example 1 Find the work required to compress a spring from its natural length of 50 cm
to a length of 40 cm if the spring constant is 20 N/m.
Solution It is often easiest to convert units into meters, so that we can be certain our units
will work out properly. In this case we want to compress the spring by an amount of 10 cm,
which is the same as 0.1 m. Thus, we will be integrating from 0 (no compression) to 0.1 (a
compression of 10 cm). We find that
Z 0.1
Z 0.1
0.1
W =
kxdx = 20
xdx = 10x2 = 0.1 J
0
0
0
Example 2 Suppose a spring has a natural length of 1 m. A 24 Newton weight is placed
on the end of the spring, which stretches it to a length of 1.8 m.
i. Find the spring constant k.
ii. How much work is required to stretch the spring 2m beyond its natural length?
iii. How far will a 45 N weight stretch the spring?
Solution Since we know F = kd, and we have a force of 24 Newtons, with d = 0.8, we find
that
d
24
k=
=
= 30 N/m
F
0.8
Next, we apply Hooke’s law and note that to stretch the spring 2 meters we will go from
x = 0 to x = 2, so
Z 2
2
2
W = 30
xdx = 15x = 60 J
0
0
Finally, we use the relationship F = kd for F = 45 N to find that
d=
F
45
=
= 1.5 m
k
30
which means that this weight will stretch the spring to a length of 2.5 meters.
Example 3A 3 kg bucket is lifted from the ground into the air by pulling in 6 meters of
rope at a constant speed. The rope weighs 0.12 kg/m. How much work is required to lift
the bucket and rope?
Solution Since the weight of the bucket does not change, the work required to lift it is
simply force times distance, which in this case is weight times distance. In order to find the
weight of the object, we need to multiply its mass by the acceleration due to gravity, 9.8
m/s2 .
3 kg · 9.8 m/s2 · 6 m = 176.4 J
The weight of the rope, however, varies with how far down the bucket is hanging. As more
of the rope is pulled in, less of it remains hanging, and so its weight decreases. As a function
x of the height of the bucket, the weight of the rope is given by
0.12 kg/m · 9.8 m/s2 · (6 − x) m = 1.17(6 − x) N
Finally, we integrate from 0 to 6, finding that
Z 6
h
x2 6
Wrope =
1.17(6 − x)dx = 1.17 6x − = 1.17 36 − 18 = 21.06 J
2 0
0
Thus, the total work done is
176.4 + 21.06 = 197.46
Using calculus we can also calculate the work required to pump liquid up out of a given container. The method we use is to think about dividing the liquid into a number of cylindrical
slabs, treat the slabs as a solid object with a given weight, and then look in the limit as the
height of the slabs approaches zero. Let’s suppose we have a liquid with a given weight per
unit volume, ρ. The volume of a given slab will be given by
A(y)dy
where A(y) is the area of a typical cross-section. Thus, the weight of a single slab will be
given by
ρA(y)dy
which is equal to the force required to lift the slab. Finally, we need to calculate how much
distance we need to lift the slab. If the top of the container is at some height h, then we will
need to lift the slab a distance
(h − y)
where y is the height of the slab itself. Summing up the work required to lift each individual
slab leads to the integral
Z
W = ρA(y)(h − y)dy
Example 4 Suppose a cylindrical tank of water has a radius of 1 meter, and a height of 10
meters. Further, suppose that this tank is filled to the brim with water. Use the fact that
water has a density of 1000 kg/m3 to determine how much work is required to pump all of
the water up out of the tank.
Solution The first thing to do is find the weight of the water, which we do by multiplying
by 9.8, the constant of acceleration due to gravity on earth at sea level. We then find
9.8 · 1000 = 980 N/m3
The volume of a typical slab is given by
π22 dy = 4πdy m3
and the work required to lift a slab is equal to its weight, so we find
F (y) = 3920π N
Finally, we need to lift a slab from a height y to a height 10, so we need to lift each slab a
distance (10 − y). Integrating over the limits 0 to 10 (the portions of the tank where water
is present - all of it), we find the total work as
Z 10
10
W =
3920π(10 − y)dy = 3920π(10y − y 2 /2) = 196000π ≈ 615752.16 N
0
0
Example 5 Consider the cone generated by rotating the line x = y/2 around the y-axis.
Suppose that this forms a tank with height 4m, and that the tank is filled up to 3m with
olive oil, weighing 8953.98544 Newtons per meter cubed. How much work is required to
pump the liquid to the rim of the tank?
SolutionWe begin by finding the volume area of a cylindrical slab, given by
A(y)dy = π(y/2)2 =
π 2
y dy m3
4
The force required to lift this slab is given by its weight
8953.98544 ·
π 2
y dy = 2238.49636πy 2 dy N
4
The distance through which F (y) must act to lift this slab to the level of the rim is 4 − y,
so integrating from 0 to 3 (the portion of the tank containing liquid), we find that
Z 3
4
y 4 3
W =
2238.49636πy 2 (4 − y)dy = 2238.49636π( y 3 − ) = 110760.989 J
3
4 0
0