Chapter 6
Work, Kinetic Energy and Potential
Energy
6.1
The Important Stuf
6.1.1 Kinetic Energy
For an object with mass
m
and speed
v
, the
kinetic energy
is defned as
K
=
1
2
mv
2
(6.1)
Kinetic energy is a scalar (it has magnitude but no direction); it is always a positive
number; and it has SI units o± kg
·
m
2
/
s
2
. This new combination o± the basic SI units is
known as the
joule
:
1joule = 1J = 1
kg
·
m
2
s
2
(6.2)
As we will see, the joule is also the unit o± work
W
and potential energy
U
. Other energy
units o±ten seen are:
1erg = 1
g
·
cm
2
s
2
= 10

7
J
1eV = 1
.
60
×
10

19
J
6.1.2 Work
When an object moves while a ±orce is being exerted on it, then
work
is being done on the
object by the ±orce.
I± an object moves through a displacement
d
while a
constant
±orce
F
is acting on it, the
±orce does an amount o± work equal to
W
=
F
·
d
=
Fd
cos
φ
(6.3)
where
φ
is the angle between
d
and
F
.
Work is also a scalar and has units o± 1N
·
m. But we can see that this is the same as
the joule, defned in Eq. 6.2.
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CHAPTER 6. WORK, KINETIC ENERGY AND POTENTIAL ENERGY
Work can be negative; this happens when the angle between force and displacement is
larger than 90
◦
. It can also be
zero
; this happens if
φ
= 90
◦
. To do work, the force must
have a component along (or opposite to) the direction of the motion.
If several diFerent (constant) forces act on a mass while it moves though a displacement
d
, then we can talk about the
net work
done by the forces,
W
net
=
F
1
·
d
+
F
1
·
d
+
F
1
·
d
+
...
(6.4)
=
p
s
F
P
·
d
(6.5)
=
F
net
·
d
(6.6)
If the force which acts on the object is
not
constant while the object moves then we must
perform an integral (a sum) to ±nd the work done.
Suppose the object moves along a straight line (say, along the
x
axis, from
x
i
to
x
f
) while
a force whose
x
component is
F
x
(
x
) acts on it. (That is, we know the force
F
x
as a function
of
x
.) Then the work done is
W
=
i
x
f
x
i
F
x
(
x
)
dx
(6.7)
²inally, we can give the most general expression for the work done by a force. If an object
moves from
r
i
=
x
i
i
+
y
i
j
+
z
i
k
to
r
f
=
x
f
i
+
y
f
j
+
z
f
k
while a force
F
(
r
) acts on it the work
done is:
W
=
i
x
f
x
i
F
x
(
r
)
dx
+
i
y
f
y
i
F
y
(
r
)
dy
+
i
z
f
z
i
F
z
(
r
)
dz
(6.8)
where the integrals are calculated along the path of the object’s motion. This expression
can be abbreviated as
W
=
i
r
f
r
i
F
·
d
r
.
(6.9)
This is rather abstract! But most of the problems where we need to calculate the work done
by a force will just involve Eqs. 6.3 or 6.7
We’re familiar with the force of gravity; gravity does work on objects which move ver
tically. One can show that if the height of an object has changed by an amount Δ
y
then
gravity has done an amount of work equal to
W
grav
=

mg
Δ
y
(6.10)
regardless of the horizontal displacement. Note the minus sign here; if the object increases
in height it has moved
oppositely
to the force of gravity.
6.1.3 Spring Force
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