Download 08-1 Note 08 Work and Kinetic Energy

Note 08
Work and Kinetic Energy
Sections Covered in the Text: Chapters 10 & 11, except 11.7. Chapter 12 read only.
In this note we begin the study of mechanical energy
and the ways in which mechanical energy can be
transferred from one system to another. Energy is
another word used in everyday speech that has a very
specific and subtly different definition in physics. We
shall see that certain problems in physics can be
solved more easily by considering them from the
point of view of mechanical energy than by applying
Newton’s laws directly. We examine the concept of
work and the relationship between work and
mechanical energy. We begin with the idea of a
system and its environment.
A System and its Environment
In physics it is useful to envoke the concept of a system
and its environment. A system is usually an object
under study, while an environment is what lies
outside a system and what is therefore irrelevant to
the problem. In any problem it is useful to identify the
system clearly so as not to waste time with effects that
have no bearing on the question being asked.
For example, in Figure 8-1 is shown two views of an
object falling toward the Earth. In (a) is shown the
object itself and the force exerted on it: the force of
gravity. In (b) is shown a wider perspective, including
the Earth, the source of the force. The force the Earth
exerts on the object is shown as well as the force the
object exerts on the Earth (via the third law). For a
certain class of problems the system is just the object
itself, modelled as a particle. Even though the object is
falling toward the Earth, the Earth plays no role in the
problem other than to provide the force on the object.
object
object
m
But in other types of problems, and especially those
pursued from the point of view of energy, the system
must be taken as is shown in Figure 8-1b; the Earth
must be included. This idea may not be clear at the
moment, but we shall enlarge upon it presently in an
effort to make it clearer.
Work Done by the Agent of a Constant
Force
Definition of Work
We all have a notion of the meaning of work as we use
the word everyday. We say that the mastery of
physics takes a lot of “work”. Perhaps you have had
the experience of pushing on an object to try to make
it move when it just would not move. Even if the
object did not move you might claim to have done a
lot of “work”, especially if the effort produced a good
sweat.
In physics, work is defined very precisely and in a
way that is subtly different from its usage in everyday
speech. If you push on an object and the object does
not move, then according to its definition in physics,
you do no work! The work done is non-zero only if
the object on which the force is exerted moves
through some displacement, as we shall see in the
next section.
Work Done by the Agent of a Constant Force
We first define work in the simplest situation, when
the force applied is constant—constant in the sense of
being independent of time or position. Consider a
particle subject to a constant force F moving through
a displacement Δr along a straight line that makes an
angle θ with Δr (Figure 8-2).
m
earth
(a)
Me
(b)
Figure 8-1. An effort to illustrate what is meant by a system
and its environment. The Earth is part of the environment
in (a), whereas in (b) it is part of the system.
Figure 8-2. A constant force F is shown moving an object
through some displacement. The agent of the force does an
amount of work.
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Note 08
The work done on an object by the agent of a force can
be defined in these words:1
The work W done by an agent exerting a constant force
of magnitude F on a particle is the product of the
component Fcos θ of the force along the direction of the
displacement of the point of application of the force and
the magnitude ∆r of the displacement:
W = FΔr cosθ .
In Figure 8-3a, F is perpendicular to the displacement
and therefore does no work (θ = π/2 and cosθ = 0). In
(b) and (d) the work done by F is negative (cos θ is
–ve since θ > π/2). Only in (c) is the work done by F
positive. Let us consider an example.
Example Problem 8-1
Calculating Work in a Simple Case
…[8-1]
Work is a scalar with dimensions M.L2.s–2 and units
newton.meter (N.m). 1 newton.meter is given the
name joule (abbreviated J) in honor of a 19th century
physicist. Thus according to this definition if ∆r = 0,
then W = 0, consistent with our earlier statement.
But work defined by eq[8-1] depends on θ as well as
on ∆r. Depending on the value of θ, work can be
positive, negative or zero. To illustrate this, consider
four cases of an object moving to the right (Figure 83). In each case a force of the same magnitude F is
applied to the object, but in different directions. Is the
work done by the agent of the force the same in all
cases?
A constant force of 1.50 N is applied at an angle of 60˚
above the horizontal to a 2.00 kg block in contact with
a horizontal frictionless surface (Figure 8-4). The block
moves a distance of 0.500 m along the surface.
Calculate the work done by the agent of the force
during this movement.
1.50 N
60˚
2.00 kg
Figure 8-4. A force applied to an object.
Solution:
Applying the definition of work done by the agent of
a constant force, eq[8-1], we have
W = FΔr cosθ = (1.50N)(0.500m)cos60 o
= 0.375 J.
It is useful to note that the work done is independent
of the mass of the object.
The most elegant definition of work uses the scalar
dot product (discussed briefly in Note 02). In
preparation we digress with a little vector analysis.
Math Unit 6: Vector Analysis
Figure 8-3. Four cases when a force of magnitude F is
applied to an object in different directions. In each case the
object is assumed to be moving to the right.
1
The work done is often referred to as the work done by the
force. But the work is done by the source or agent of the force not the
force itself. And the agent of the force is the environment. We shall
assume this if the phrase “work done by the force” appears in these
notes from now on.
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Using the scalar (dot) product we can frame a more
compact and general definition of work than is given
by eq[8-1].
Scalar Product of Two Vectors
As described in Note 02, the dot product of any two
vectors A and B is the scalar quantity
A • B ≡ AB cosθ ,
…[8-2]
Note 08
where A and B are the magnitudes of the two vectors
and θ is the smaller of the two angles between them
(Figure 8-5).
Solution:
We shall use the definition of the vector dot product.
First note that the magnitude of A is
A = Ax2 + Ay2 + Az2 = 1+ 4 + 9 = 14 ,
B = 29 .
and of B:
€
Figure 8-5. Two vectors.
It is useful to express the vectors in terms of unit
vectors. Recall that the concept of a unit vector was
introduced in Note 02 for the purpose of indicating
direction. We shall see that the following relationships
between the dot products of the unit vectors will
prove useful:
) ) ) ) ) )
i • i = j • j = k • k = 1,
Therefore from eqs[8-2] and [8-3]:
€ A• B
cos θ =
AB
=
Ax Bx + Ay By + Az Bz
= 0.992583 .
AB
Thus θ = ArcCos(0.992583) = 7˚. The smaller of the
two angles between the two vectors is 7 ˚.
) ) ) ) ) )
i • j =i •k = j •k = 0.
Using the vector dot product we can rewrite the
definition eq[8-1] in the form:
Thus any two vectors in 3D space can be expressed in
component form as
W = F • Δr .
and
)
)
)
A = Axi + Ay j + Azk and
)
)
)
B = Bx i + By j + Bzk .
Thus the scalar product of these vectors is
A • B = Ax Bx + Ay By + Az Bz .
…[8-3]
We also have for the dot product of a vector with itself
the result
A • A = Ax2 + Ay2 + Az2 = A 2 .
Let us consider an example.
Example Problem 8-2
Finding the Angle Between Two Vectors
Find the smallest angle between the vectors
)
)
)
A = 1.00i + 2.00 j + 3.00k
and
)
)
)
B = 2.00i + 3.00 j + 4.00k .
…[8-4]
We repeat, eq[8-4] applies only if the force is constant .
If the force is variable in the sense of depending on
time or position then the work done must be found
using calculus. This more general case we examine
next.
Work Done by the Agent of a Varying Force
Work can be done by a force that varies with time,
position or both. We begin by focussing our attention
on a force that depends only on x.
Consider a particle being moved along the x-axis
from xi to xf by a force that depends in some way on x.
In general, this force may be in any direction. Here we
consider only the x-component of the force. The xcomponent might have the form shown in Figure 8-6a.
To recycle the idea of a constant force and to utilize
eq[8-4] we imagine the x-domain to be divided into a
large number of equal intervals. Let us focus our
attention on an arbitrary (nth) interval; call it ∆x.
If the interval is narrow enough, then the x-component of force that applies over that interval is nearly a
constant with some average value, say F x. The work
done in that interval is the product:
Wn ≈ Fx Δx .
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Note 08
Now the area under the curve bounded by ∆x is just
the width times the average height, or F x∆x. Thus the
work done in the nth interval equals the area that
interval occupies under the force curve (Figure 8-6a).
We can find an approximate value for the total work
done over the whole range by adding up the
contributions of work done over all of these intervals.
The result is:
xf
xf
W = lim Wapprox = lim ∑ Fx Δx = ∫x Fx dx .
Δx→ 0
Thus
Δ x →0
i
xi
xf
W = ∫x Fx dx ,
…[8-6a]
i
or, written in function form
xf
W = ∫x F(x)dx .
…[8-6b]
i
According to calculus, this expression exactly equals
the area under the force curve between the limits xi
and xf (Figure 8-6b).
For most purposes in a first course in physics, eq[86b] is sufficient for calculating the work done by a
varying force. However, our treatment takes account
of just one force. If a number of forces act on the object
simultaneously then the expression for the force in
eq[8-6] must be replaced by a sum of forces:
xf
Wnet = ∫x
i
(∑ F )dx .
x
And finally, the most general form of the work done
allows for multiple forces acting in different directions (a situation likely to occur only in an advanced
course in physics). This general expression involves
the vector dot product:
r2
Wnet = ∫r
1
Figure 8-6. How the work done by a force that varies with x
is found using calculus. The total work done between the
two positions xi and xf is equal to the area under the force
curve between those positions.
xf
Wapprox = ∑ Fx Δx .
…[8-5]
xi
This is only an approximation because the ∆xs are of
finite size. To increase the accuracy we can take the
limit of this expression as we make the width of the
intervals vanishingly small, i.e., as ∆x → 0. The result
is
08-4
(∑ F) • dr .
…[8-7]
A good example of a single varying force is the force
exerted by a spring. This case we consider in the next
section.
Work Done by a Spring
A classic example of a force that varies in a simple
way with displacement is the force exerted by a
spring on a mass connected to its end (Figure 8-7a).
For convenience we assume that the mass rests on a
horizontal frictionless surface. Robert Hooke (a
contemporary of Isaac Newton) found experimentally
that if the mass is displaced a small amount x from
equilibrium (the position x = 0), then the spring exerts
a force on the mass given by
Fs (x) = –kx .
…[8-8]
k is a constant called the force constant of the spring
and has units N.m–1. k is set by the construction of the
Note 08
spring, being large for a “hard” spring and small for a
“soft” spring.2
The negative sign in eq[8-8] indicates that the force
is a restoring force—restoring in the sense that it tends
to restore the mass to the system’s equilibrium
position. If the mass is displaced to the right of the
equilibrium position, then the force exerted by the
spring is directed to the left. If the mass is displaced to
the left of the equilibrium position, then the force
exerted by the spring is directed to the right. Over
small displacements, the variation of the force is
linear, varying with x as sketched in Figure 8-7b.
the area under the force curve between x = –xmax and x
= 0. It is positive since the force and the displacement
are in the same direction.
This procedure can be applied to calculate the work
done by the spring when the mass moves between
any two positions, say xi and xf. The result is
xf
Ws =
1
∫ (–kx)dx = 2 kx
2
i
1
2
– kx 2f .
xi
…[8-9]
Clearly, in a movement of the mass from x = – xmax to x
= +x max the work done by the spring is zero. In the
segment between x = –xmax to x = 0, the spring does a
positive amount of work. In the segment between x =
0 and x = +x max the spring does a negative amount of
work. Let us consider a numerical example.
Example Problem 8-4
Work Done by a Spring
A mass is connected to the end of a spring in an
arrangement as shown in Figure 8-7a. The spring
constant is k = 0.020 N.m–1. The mass is released from
a position x i = 0.25 m. How much work is done by the
spring when the mass moves to the position xf = – 0.10
m?
Solution:
Using eq[8-9] we have
1
2
W = k(xi2 – x 2f )
Figure 8-7. In (a) is shown a mass m attached to the end of a
spring. In (b) is shown a sketch of how the spring force
varies with x. The variation is linear within the spring’s
elastic limit.
We can calculate the work done by the spring on the
mass over any movement of the mass. Let us suppose
that the mass is released from a position of x = –xmax.
We can calculate the work done by the spring force as
the mass moves from this position to the equilibrium
position x = 0. According to eq[8-6] we have
xf
Ws =
∫
xi
0
Fs (x)dx =
∫
– xmax
(–kx)dx =
1
2
kx2max .
As can be seen from Figure 8-7b this result is equal to
=
1
(0.020N.m–1 )[(0.25m) 2 – (–0.10m)2 ]
2
= 5.25 x 10–4 J.
The spring does a positive amount of work.
In physics, the work done on an object in some process is interpreted as the energy acquired by the
object.3 This relationship is laid out in the so-called
work-energy theorem. An object can possess energy by
virtue of its motion. This kind of energy is called
kinetic energy, as we shall see next. (In Note 09 we
shall see that an object can also possess energy by
virtue of its position relative to some reference
3
2
We emphasize for completeness that eq[8-8] applies only if the
spring is not deformed or stretched beyond its elastic limit.
And we are talking here about mechanical energy. The subject
of the internal energy of an object will be discussed in the second
half of this course.
08-5
Note 08
position.)
difference between two terms of the form
Kinetic Energy and the Work-Kinetic Energy
Theorem
In this section we introduce the idea of kinetic energy.
Consider an object being moved in a straight line over
a frictionless horizontal surface by a force or sum of
forces ΣF that is constant (Figure 8-8). For convenience
we assume that the force and displacement are in the
same direction. In a certain elapsed time ∆t the object
moves through a displacement ∆x. Since the force is
constant the object accelerates continuously and so its
speed increases from, say, vi to vf.
Figure 8-8 An object is moved by the agent of a constant
force and undergoes an acceleration and an increase in
speed.
We have seen that the work done on the object by the
agent of the constant net force is given by
xf
Wnet = ∫x
i
∑ Fdx .
Substituting ΣF = ma and using the chain rule in
calculus we have
xf
xf
i
i
Wnet = ∫x madx = ∫x m
xf
dv
dv dx
dx = ∫x m
dx
i
dt
dx dt
xf
K=
1
2
mv2 .
…[8-11]
We can therefore write
Wnet = K f – Ki = ΔK .
…[8-12]
K involves the square of the speed of the object and is
therefore called kinetic energy. Kinetic energy is therefore the mechanical energy a body possesses by virtue
of its motion. The unit of kinetic energy is the same as
the unit of work, namely joule (J). Thus the workkinetic energy theorem states in so many words that
the work done on an object (in a situation as is shown
in Figure 8-8) equals the increase in the object’s kinetic
energy.
This result enables any number of physics problems
to be solved in terms of mechanical energy rather than
via the kinematic equations of motion. Sometimes the
energy approach leads to a solution more quickly,
other times the kinematic equations are faster. It
depends on the problem. Let’s consider an example.
Example Problem 8-4
Finding the Speed of an Object
A block of mass 2.00 kg in a situation as shown in
Figure 8-8 is initially at rest on a horizontal frictionless
surface. A force of 2.00 N is applied to the block in the
direction shown in the figure for 5.00 s. Calculate the
final speed of the block.
Solution:
Applying the second law to the block, the block’s
acceleration is
a=
F (2.00N)
=
= 1.00 m.s–2.
m (2.00kg)
The distance travelled in 5.00 s is
= ∫x mvdv .
Δx = ut + at 2 = 0 + (1.00m.s –2 )(5.00s)2
This amounts to a change in variable from x to v. Thus
= 12.5m .
1
2
i
1
2
Wnet = mv 2f –
1
2
mvi2 .
…[8-10]
Since vf > vi, Wnet is positive. The work done equals the
08-6
1
2
Applying the work-energy theorem, the work done
equals the increase in the block’s kinetic energy:
Note 08
1
2
uct of the net applied force vector and the velocity
vector. This result is useful in the solving of certain
problems.
FΔx = mv 2
or
1
2
(2.00N)(12.5m) = (2.00kg)v2
Example Problem 8-5
Average Power
v = 5.00 m.s–1.
Solving for v:
It should be obvious this problem could have been
solved just as well by kinematics alone.
Power
In physics, we are often interested in the rate at which
energy changes, or is transferred from one system to
another, as much as in the energy itself. The rate at
which work is done or energy is transferred is called
power. Two types of power are defined: average power
and instantaneous power.
Average Power
If an amount of work ∆W is done in an elapsed time
∆t then the average power is defined as the work done
divided by the elapsed time:
P=
ΔW
.
Δt
The units of average power are joules per second (j.s ).
1 J.s –1 is given the special unit Watt (abbreviated W).
Instantaneous Power
The instantaneous power is defined as the limit of the
average power as ∆t → 0:
Δt →0
Δt → 0
ΔW dW
=
.
Δt
dt
…[8-14]
The unit of instantaneous power is also J.s–1 or Watt.
Instantaneous power can also be expressed in terms of
force and velocity. Substituting the definition of work,
eq[8-7], into eq[8-14] we have
P=
dW
dr
=F•
= F•v
dt
dt
P=
ΔW 2.00(J)
=
= 4.00 Watts.
Δt
0.500(s)
In the elapsed time given, energy is being transferred
from the agent of the force to the object at the average
rate of 4.00 joules per second.
Example Problem 8-6
Instantaneous Power
In the example shown in Figure 8-4 what is the
instantaneous power at a clock time of 2.00 s?
Solution:
We intend to apply the expression given in eq[8-15b],
namely
P = Fv cos θ .
To do this we need to find the velocity at the clock
time of 2.00 s. Using kinematics we have
v = 0 + at =
Thus
…[8-15b]
(1.50N)cos60o
(2.00s) m.s –1
(2.00kg)
= 0.75 m.s–1.
…[8-15a]
or since this is a vector dot product,
P = Fv cos θ ,
Solution:
According to the definition of average power, eq[813],
…[8-13]
–1
P = lim P = lim
The source of a force applied to an object does 2.00 J of
work in an elapsed time of 0.500 s. Calculate the
average power.
P = (1.50N)(0.75m.s –1 )cos60o = 0.563 W.
At the precise clock time of 2.00 s, energy is being
transferred at the rate of 0.563 joules per second.
where θ is the smallest angle between F and v . Thus
the instantaneous power equals the vector dot prod-
08-7
Note 08
To Be Mastered
•
•
•
•
•
Definitions: system, environment
Definitions: work done by a constant force, work done by a varying force
Definitions: unit vector, vector dot product
Definitions: work-kinetic energy theorem, kinetic energy
Definitions: average power, instantaneous power
Typical Quiz/Test/Exam Questions
1.
(a) State Hooke’s Law.
(b) Define the spring constant of an elastic spring.
(c) Explain how a stretched spring can be used to measure force.
2.
(a) Define what is meant by the work done on an object.
(b) Give an example of the relationship between work and one type of mechanical energy. Illustrate your
answer by means of a sketch showing all important quantities.
3.
(a) Define what is meant by the work done by a constant force on an object.
(b) Write down an expression for the work done by a force F(x) between xi and xf.
4.
A block of mass 1.0 kg is in contact with a horizontal surface (see the figure). The coefficient of kinetic friction
between the block and the surface is 0.6 N.kg–1. A constant force of 2.0 N is applied to the block in the
direction shown. When the block moves a distance of 1.0 m along the surface how much work has been done
by the agent of the force?
2.0 N
60˚
1.0 kg
1.0 m
5.
A block of mass 1.0 kg is in contact with an inclined plane (see the figure). The coefficient of kinetic friction
between the block and the plane is 0.6 N.kg–1. A constant force of 2.0 N is applied to the block in the direction
shown. When the block moves a distance of 1.0 m up the incline how much work has been done by the agent
of the force?
08-8
Note 08
2.0 N
1.0 kg
1.0 m
30˚
08-9