Work
Definition 1.
The work done by a force F (x) in moving an object from x = a to x = b is
W =
Z
b
F (x) dx
a
In particular, if the force is a constant, F , independent of x, the work is F (b − a).
In SI1 metric units the force F has units newtons (which are kilograms meters per second
squared), x has units meters and the work W has units joules (which are newton meters or
kilograms meters squared per second squared).
Here is some motivation for this definition. Consider a particle of mass m moving along
the x–axis, that starts at a at time α and ends at b at time β. Suppose that at time t, the
particle is at x(t) and that, when it is at position x it is subjected to a force F (x). Then
Newton’s law of motion says that
m
d2 x
(t)
=
F
x(t)
dt2
Multiply this equation by the particle’s velocity v(t) =
m
dx
(t)
dt
at time t.
dx
d2 x dx
(t)
(t)
=
F
x(t)
(t)
dt2
dt
dt
(1)
Now the left hand side
d2 x dx
d 1 dx 2
d 1
2
m 2 (t) (t) =
m
(t)
=
mv(t)
dt
dt
dt 2
dt
dt 2
is exactly the derivative of the kinetic energy 12 mv(t)2 . So when we integrate (1) from t = α
to t = β, we get, by the fundamental theorem of calculus,
Z β
dx
1
1
2
2
m v(β) − m v(α) =
(t) dt
F x(t)
2
2
dt
α
Finally, on the right hand side, make the substitution x = x(t), dx =
1
1
m v(β)2 − m v(α)2 =
2
2
Z
dx
(t) dt.
dt
b
F (x) dx
(2)
a
The work W of Definition 1 is the change in kinetic energy from the time the particle was
at x = a to the time it was at x = b.
1
SI is short for “le système international d’unités” which is French for “the international system of units”.
c Joel Feldman. 2014. All rights reserved.
1
January 28, 2014
Example 2 (Escape Velocity)
The gravitational force of the Earth acting on a particle of mass m at a height x above the
GM m
surface of the Earth is (R+x)
2 , where G is the gravitational constant, M is the mass of the
Earth and R is the radius of the Earth. So the work required to lift a particle from the
surface of the Earth to a height h is
h
Z h
GMm GMm
GMm GMm
=
dx = −
−
2
R+x 0
R
R+h
0 (R + x)
1
(You can find the indefinite integral of (R+x)
2 by making the substitution u = R + x and
R dx
1
1
then integrating u2 . But it is easier to just guess that (R+x)
2 = − R+x + C and verify that
1
1
d
− R+x
is indeed (R+x)
the guess is correct by checking that dx
2 .) In particular to lift the
GM m
particle all the way to h = ∞ takes work R . If all of this energy is to come from the
particle’s initial kinetic energy 21 mv 2 , i.e. the particle is not equipped with a rocket engine,
we need that
r
1 2 GMm
2GM
mv ≥
⇐⇒ v ≥
2
R
R
q
The right hand side, 2GM
, is called the escape velocity of the Earth.
R
Example 2
Example 3 (Hooke’s Law)
Hooke’s Law says that when a (linear) spring is stretched/compressed by x units beyond its
natural length (with x > 0 for stretching and x < 0 for compression), it exerts a force of kx,
where the constant k is the spring constant of that spring. So to stretch a spring by L units
from its natural length takes a work of
Z L
1
W =
kx dx = kL2
2
0
Example 3
Example 4 (Pumping Out a Reservoir)
A cylindrical reservoir of height h and radius r is filled with a fluid of density ρ. (The lengths
h and r could have units meters and the density ρ could have units kilograms per cubic
meter.) We would like to know how much work is required to empty the reservoir.
x
r
h
c Joel Feldman. 2014. All rights reserved.
2
dx
x
January 28, 2014
Slice the reservoir into thin, horizontal, cylindrical pancakes, as in the figure above. The
pancake at height x above the bottom has
•
•
•
•
thickness dx and
a cross–section which is a circular disk of radius r and area πr 2 .
So the pancake has volume πr 2 dx and
mass ρ × πr 2 dx.
Near the surface of the Earth gravity exerts a force of mg on a body of mass m. So the
pancake, which has mass πρr 2 dx, feels a gravitational force of πρgr 2 dx. To remove this
pancake from the reservoir we need to lift it a height of h − x working against gravity, which
takes work πρgr 2 (h − x) dx. The total work to empty the whole reservoir is
h
h
ih π
1
2
= ρg r 2 h2
W =
π ρg r (h − x) dx = π ρg r
(h − x) dx = π ρg r − (h − x)
2
2
0
0
0
h
i
R
d
− 12 (h−x)2 = h−x
The integral (h−x) dx = − 21 (h−x)2 +C was just guessed. Because dx
the guess is correct. The integral can also be done by substituting u = h − x, du = −dx,
Rh
R0
Rh
2
giving 0 (h − x) dx = − h u du = 0 u du = h2 .
Z
2
2
Z
2
h
Example 4
c Joel Feldman. 2014. All rights reserved.
3
January 28, 2014