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Chapter 6: Work & Energy
THE COURSE THEME is
NEWTON’S LAWS OF MOTION!
• So Far: Motion analysis with forces.
• NOW : An alternative analysis using
the concepts of Work & Energy.
– Easier? My opinion is yes!
Conservation of Energy is
NOT a new law!
– We’ll see that this is just Newton’s Laws of
Motion re-formulated or re-expressed (translated)
from Force Language to Energy Language.
Newton’s
nd
2
Law:
• So far, we’ve expressed Newton’s Laws of
Motion using the concepts of position,
displacement, velocity, acceleration & force.
• Newton’s Laws with Forces: Are quite general
(macroscopic objects). In principle, they could be
used to solve any dynamics problem, But, often, they
are very difficult to apply, especially to very
complicated systems. So, alternate formulations have
been developed. These alternate formulations are
often easier to apply.
• Also, a practical problem is that, often, we may not
even know all of the forces.
Newton’s
nd
2
Law:
• One alternate approach uses Energy
instead of Force as the most basic
physical quantity.
• So, we’ll discuss Newton’s Laws in
Energy language.
• Before we discuss these, we need to
learn some vocabulary in
Energy Language.
nd
2
Newton’s
Law:
• Energy: A very common term in
everyday usage. Everyday meanings
might not coincide with the
PHYSICS meaning!
• Every physical process involves
energy or energy transfer or
transformations. Energy in physics
can be somewhat abstract.
Work
• A connection between force & energy
is work. Work depends on the force, the
displacement & the direction between them.
• Work in physics has a more specific
meaning than in everyday usage.
• Work: Precisely defined in physics.
Describes what is accomplished by
a force in moving an object through
a distance.
Work Done by Constant Force
• Work: Precisely defined in physics. Describes
what is accomplished by a force in moving an
object through a distance.
For an object moving under
the influence of a Constant
Force, (as in the figure!)
d
Work done (W)  product
of the magnitude of the
W

F
d
=
Fd
cosθ
||
displacement (d)  the
component of force parallel
to the displacement (F|| ).
Work: W  FΔr
•For 1-dimensional
motion, W = F Δx
• In 2- or 3-dimensions, the vector nature
of the force F & displacement Δr must
obviously be taken into into account:
W = FΔrcos θ
• θ is the angle between the force and the
displacement
Work and Directions
•The component of F along the
displacement direction is:
Fcos θ  F||
•When F|| is parallel to the
displacement, the work is positive.
•When F|| is anti- parallel to the
displacement, the work is negative.
•When the component of the
force is perpendicular to the
displacement, the work is zero.
Work
• Experiments have verified that although various
forces produce different times & distances, the
product of the force & the distance remains the same.
• To accelerate an object to a specific velocity, you
can exert a large force over a short distance or a
small force over a long distance
Work, Force and Displacement
Work (W) is done by a force
acting on an object.
• W depends on the force F acting on the object
and on the object’s displacement d.
• The value of W depends on the direction of the
force relative to d.
• W may be positive, negative, or zero, depending
on the angle θ between the force and the
displacement.
• If the displacement is zero (the object does not
move), then W = 0, even though the force may be
very large.
What Does the Work?
• When an “agent” (another mass from N’s
3rd Law!) applies a force to an object and
does an amount of work W on that object,
the object will do an amount of work equal
to –W back on the agent
– The forces form a Newton’s 3rd Law
action-reaction pair.
• Multiple agents (masses):
– When multiple agents act on an object, you
can calculate the work done by each
separate agent
Work and Amplifying Force
• See figure. Suppose the
person lifts his end of the
rope through a distance L.
• The pulley will move
through a distance (½ )L
• Work on the crate:
Won crate = (2T)(L/2) = TL
• Work on the rope:
Won rope = TL
• Work done on the rope is equal to the work done
on the crate
Work and Amplifying Force
• Work on the crate:
Won crate = (2T)(L/2) = TL
• Work on the rope:
Won rope = TL
• Work done by the person is effectively
“transferred” to the crate.
• Forces can be amplified, but work can’t be increased this way.
The force is amplified, but not the work
• The associated displacement is decreased
• The work-energy theorem suggests work can be converted to
energy, but since work cannot be amplified the exchange will
not increase the amount of energy available
• The result that work cannot be amplified is a consequence of the
principle of conservation of energy
Work Done by a Constant Force
Work: W  F||d = Fd cosθ
For a CONSTANT force!
W = F||d = Fd cosθ
• Consider a simple special case when
F & d are parallel: θ = 0, cosθ = 1
 W = Fd
• Example: d = 50 m, F = 30 N
 W = (30N)(50m) = 1500 N m
SI Work Units
Newton - meter = Joule
1 N m = 1 Joule = 1 J
W  F||d = Fdcosθ
• Its possible to exert a
force & yet do no work!
• Could have d = 0  W = 0
• Or could have F  d
 θ = 90º, cosθ = 0
 W=0
• Example, walking at
constant speed v with a
grocery bag.
Example
W = F||d = Fd cosθ
m = 50 kg, FP = 100 N, Ffr = 50 N, θ = 37º
Example
W = F||d = Fd cosθ
m = 50 kg, FP = 100 N, Ffr = 50 N, θ = 37º
Solution:
WG = mgxcos(90) = 0, WN = FN xcos(90) = 0
WP = Fpxcos(37) = (100)(40)cos(37) = 3200 J
WFfr = Ffrxcos(180) = - 2000J
Wnet = 0 + 0 + 3200 -2000 = 1200 J
Solving Work Problems
1. Sketch a free-body diagram.
2. Choose a coordinate system.
3. Apply Newton’s Laws to determine any
unknown forces.
4. Find the work done by a specific force.
5. Find the net work by either
a. Find the net force & then find the work
it does, or
b. Find the work done by each force & add.
W = F||d = Fd cosθ
Typical Problem
An object is displaced by
a force F on a frictionless,
d
horizontal surface. The
free body diagram is
Angles for forces in the figure:
shown. The normal force
Normal Force:
FN & weight mg do no
θ = 90°, cosθ = 0
work in the process, since
Weight:
both are perpendicular to θ = 270° (or - 90°),
cosθ = 0
the displacement.
Example: Work on a Backpack
(a) Calculate the work a hiker must do on
a backpack of mass m = 15 kg to carry it
up a hill of height h = 10 m, as shown.
(b) Calculate the work done by gravity on
the backpack.
(c) Calculate the net work done on the
backpack. For simplicity, assume that the
motion is smooth & at constant velocity
(zero acceleration).
For the hiker:
∑Fy = 0 = FH – mg
 FH = mg
WH = FHdcosθ = FHh
Conceptual Example:
Does Earth do Work on the Moon?
The Moon revolves around
the Earth in a nearly circular
orbit, with approximately
constant tangential speed,
kept there by the gravitational
force exerted by the Earth.
Does gravity do
(a) positive work
(b) negative work, or
(c) no work at all on
the Moon?
Example
The force shown has magnitude FP = 20 N & makes
an angle θ = 30° to the ground. Calculate the work
done by this force when the wagon is dragged a
displacement d = 100 m along the ground.
Example
The force shown has magnitude FP = 20 N & makes
an angle θ = 30° to the ground. Calculate the work
done by this force when the wagon is dragged a
displacement d = 100 m along the ground.
Solution:
WP = FPdcos(30) = (20)(100)cos(30) = 1732 J
Example
A man cleaning a floor pulls a vacuum cleaner with a force F = 50 N
at an angle θ = 30º with the horizontal. Calculate the work done by
the force F as the vacuum cleaner is moved a distance
Δr = 3.0 m to the right.
Δr = 3.0 m
Solution
A man cleaning a floor pulls a vacuum cleaner with a force F = 50 N
at an angle θ = 30º with the horizontal. Calculate the work done by
the force F as the vacuum cleaner is moved a distance Δr = 3.0 m
to the right.
W = Fd cosθ
= (50)(3)cos(30º)
= 130 J
Δr = 3.0 m
Kinetic Energy &
Work-Energy Theorem
• Energy: Traditionally defined as the
ability to do work. We now know that
not all forces are able to do work;
however, we are dealing in these
chapters with mechanical energy,
which does follow this definition.
• Kinetic Energy  The energy of motion
“Kinetic”  Greek word for motion
An object in motion has the ability to do work.
Kinetic Energy
• Find the work done on an object as a force F moves
it from an initial position xi to a final position xf
Wnet = FnetΔx = (ma)Δx (1)
• If it is a constant, the acceleration a can be expressed
in terms of velocities (1 d kinematic equations):
• Combining gives:
Wnet = (½)mvf² - (½)mvi² (2)
(½)mv²  kinetic energy  KE
• KE is the energy due to the motion of the object.
• (1) & (2) together give the Work-Energy Theorem:
Wnet = KE
The Work-Energy Theorem
Wnet = KE
Physics of the Work-Energy
Theorem:
The kinetic energy of an object
can be changed by doing work
on the object
• Consider an object moving in straight line. It starts at
speed v1. Due to the presence of a net force Fnet, (≡ ∑F),
it accelerates (uniformly) to speed v2, over a distance d.
Newton’s 2nd Law: Fnet= ma (1)
1d motion, constant a
 (v2)2 = (v1)2 + 2ad
 a = [(v2)2 - (v1)2]/(2d) (2)
Work done: Wnet = Fnet d
(3)
Combine (1), (2), (3):
Fnet= ma
a = [(v2)2 - (v1)2]/(2d)
Wnet = Fnet d
(1)
(2)
(3)
• Combine (1), (2) & (3):
 Wnet = mad = md [(v2)2 - (v1)2]/(2d)
OR
Wnet = (½)m(v2
2
) –
(½)m(v1
2
)
• Summary: The net work done by a constant force
in accelerating an object of mass m from v1 to v2 is:
 KE
DEFINITION: Kinetic Energy (KE)
(Kinetic = “motion”, units are Joules, J)
• We have: The
WORK-ENERGY Theorem
Wnet = KE
( = “change in”)
• We’ve shown this for a 1d constant force.
However, it is valid in general!
The net work done on an object =
the change in it’s KE.
Wnet = KE (1)
 The Work-Energy Theorem
Note: (1) is Newton’s
nd
2
Law in
Work & Energy language!
The net work done on an object =
the change in it’s KE: Wnet = KE
(1)
 The Work-Energy Theorem
Note:
• Wnet = work done by the net (total) force.
• Wnet is a scalar & can be positive or negative
– This is true because because KE can be both + & -.
• If Wnet is positive, the KE increases. If Wnet
is negative, the KE decreases.
• Units are Joules for both work & kinetic energy.
Note: (1) is Newton’s 2nd Law in
Work & Energy language!
• A moving hammer can
do work on a nail!
• For the hammer:
Wh = KEh = -Fd
= 0 – (½)mh(vh)2
• For the nail:
Wn = KEn = Fd
= (½)mn(vn)2 - 0
Example: Kinetic Energy & Work
Done on a Baseball
• A baseball, mass m = 145 g (0.145 kg) is
thrown so that it has a speed v = 25 m/s.
a. What is its kinetic energy?
b. What was the net work done on the ball
to make it reach this speed, starting from rest?
Example: Kinetic Energy & Work
Done on a Baseball
• A baseball, mass m = 145 g (0.145 kg) is
thrown so that it has a speed v = 25 m/s.
a. What is its kinetic energy?
b. What was the net work done on the ball
to make it reach this speed, starting from rest?
Solution:
a. KE = (½)mv2 = (½)(0.145)(25)2 = 45.3 J
b. Wnet = KE = (45.3) – 0 = 45.3 J
Work on a car to increase its kinetic energy
Calculate the net work required to accelerate a car,
mass m = 1000-kg car from v1 = 20 m/s to v2 = 30 m/s.
Work on a car to increase its kinetic energy
Calculate the net work required to accelerate a car,
mass m = 1000-kg car from v1 = 20 m/s to v2 = 30 m/s.
Solution: Wnet = KE = (½)m(v2)2 – (½)m(v1)2
Wnet = (½)(1000)[(30)2 – (20)2]
Wnet = 2.5  104 J
Conceptual Example: Work to Stop a Car
A car traveling at speed v1 = 60 km/h can brake to a
stop within a distance d = 20 m. If the car is going
twice as fast, 120 km/h, what is its stopping
distance? Assume that the maximum braking force
is approximately independent of speed.
• Definition of Work: Wnet = Fd cos (180º) = -Fd
• Work-Energy Theorem:
Wnet = KE = (½)m(v2)2 – (½)m(v1)2
(v2)2 = 0 (car stopped) so -Fd = KE = 0 - (½)m(v1)2 or
d  (v1)2
So the stopping distance is proportional to the square of the
initial speed! If the initial speed is doubled, the stopping
distance quadruples! Note: KE  (½)mv2  0 must be
positive, since m & v2 are always positive (real v).
Example
A block, mass m = 6 kg, is pulled
from rest (v0 = 0) to the right by a
constant horizontal force F = 12
N. After it has been pulled for
Δx = 3 m, find it’s final speed v.
Example
A block, mass m = 6 kg, is pulled
from rest (v0 = 0) to the right by a
constant horizontal force F = 12
N. After it has been pulled for
Δx = 3 m, find it’s final speed v.
Solution: Work-Energy Theorem
Wnet = KE  (½)[m(v)2 - m(v)2]
(1)
• If F = 12 N is the only horizontal force, we have
Wnet = FΔx
(2)
• Combine (1) & (2): FΔx = (½)[m(v)2 - 0]
• Solve for v:
(v)2 = [2Δx/m]
v = [2Δx/m]½ = 3.5 m/s