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Work & Energy
Day #1: Introduction
I. A. Define Work:
Work is a measure of what is accomplished when a
force is applied onto an object while the object
moves. The object must be moving while the force
is applied.
B. Equation of work:
Work is defined as the product of the force applied
on an object and the displacement of the object.
W  F x cos 
W  F x cos 
F = force acting
on the object
x = displacement
of the object
The angle  is the angle between the direction of the
force F and the displacement x.
C. Units for work:
The product of force and displacement gives units of
newtons times meters.
This unit is given a special name: joule = J
The relationship is given as follows:
1J 1N  m
kg  m
1J  1 2  m
s
kg  m
1J  1 2
s
2
D. Alternate expression for work:
W  F x cos 
The term F||=Fcos is the component of the force F
that is parallel to the displacement x. The work is
the product of these two parallel components.
W   F|| x
E. Work and sign (±):
W  F x cos 
W   F|| x
For the equation of work on the left, there is a
dependence on the direction of the force relative to
the displacement. If 0o ≤  < 90o, then the work is
positive. If 90o <  ≤ 180o, then the work is
negative.
If F|| points in the same direction as the
displacement x, then the work is positive.
If F|| points in the opposite direction as the
displacement x, then the work is negative.
Ex. #1: A 10.0 kg mass is pulled towards the right 10.0 m by an applied
force of 40.0 N. The applied force also points towards the right and the
object moves at a constant velocity. (a) Draw all the forces acting on
the object.
n = Normal Force
Fk = Kinetic Friction Force
Fo = Applied Force
mg = Weight
x = Motion
(b) What is the work done by the normal force?
Wnormal  n x cos   n x cos 90o  0
(c) What is the work done by the weight force of the object?
Wweight  mg x cos   mg x cos 90o  0
General Rule: Any force applied perpendicular to the
motion does not do any work!
(d) What is the work done by the applied force?
Wapplied  Fo x cos 
Wapplied  40.0 N 10.0 m  cos0
Wapplied  400 J
(e) What is the work done by the friction force?
Fk = Kinetic Friction Force
Fo = Applied Force
Since the object moves at a constant
velocity, the forces balance!
Fo  Fk  40.0 N
W friction  Fk x cos 

W friction  40.0 N 10.0 mcos 180o

  400 J
(f) What is the net work on the object?
Wnet  Wnormal  Wweight  Wapplied  W friction
Wnet  0  0  400 J   400 J 
Wnet  0
Fnet  0
Ex. #2 Is it possible to do work on an object that remains at rest? Why
of why not?
NO. Work requires a displacement:
W  F x cos 
Ex. #3: A box is being pulled across a rough floor at a constant speed.
What can you say about the work done by friction?
1) friction does no work at all
2) friction does negative work
3) friction does positive work
W  F x cos 
Friction points opposite to the
motion of the object!
W  F x cos180o  0
Ex. #4: Can friction ever do positive work?
YES!
Consider a box placed on the back of a flatbed truck:
Fs
When the truck drives forwards, the inertia of the box makes the
box appear to slide to the back of the truck.
Friction between the box and the truck pulls the box forward,
opposing its motion towards the back of the truck.
Ex. #5: In a baseball game, the catcher stops a 90-mph pitch. What
can you say about the work done by the catcher on the ball?
1) catcher has done positive work on the ball
2) catcher has done negative work on the ball
3) catcher has done zero work on the ball
The force exerted by the catcher is opposite in
direction to the displacement of the ball, so the
work is negative. Or using the definition of work
(W = F d cos  ), since  = 180o, then W < 0. Note
that because the work done on the ball is
negative, its speed decreases.
Follow-up: What about the work done by the ball on the catcher?
Ex. #6: A man with a weight of 735 N stands in an elevator that
accelerates upwards at 1.20 m/s2 for 5.00 m. (a) What is the work
done by gravity?
NOTE: Since the object moves
upwards, and the weight points
downwards,  = 180o.

Wweight  735 N 5.00 mcos 180
o

n = Normal Force
Wweight  mg x cos 
Wweight   3680 J
mg = Weight
(b) What is the work done by the normal force?
NOTE: Since the acceleration is
given, solve for the normal force
using Newton’s 2nd Law:
The sum of the forces is:
Fnet  ma
Fnet  n  mg
n  mg  ma
NOTE: the weight is
given, not the mass!
n  ma  mg
mg  735 N
735 N
m
 75.0 kg
9.80 m s 2
n  ma  mg
n  75.0 kg 1.20 m s 2   735 N
n  825 N
Wnormal  n x cos 
NOTE: Since the object moves upwards, and the normal force points
upwards,  = 0o.
Wnormal  825 N 5.00 mcos 0
Wnormal  4130 J
Ex. #7: A block is pulled towards the right with a force of 50.0 N applied
at an angle of 36.9o above the horizontal. The block is already in
motion and moves a distance of 20.0 meters at a constant velocity.
(a) What is the work done by the applied force?
Fo
n = normal
FK
 = 36.9o
m
mg = weight
Wapplied  Fo x cos 

Wapplied  50.0 N 20.0 mcos 36.9
o

 800 J
(b) What is the work done by the force of friction?
The object moves at a constant velocity, which
means the forces balance. Friction will balance the
horizontal component of the applied force.

Fk  Fo cos   50.0 N  cos 36.9

o

Wk  40.0 N 20.0 m  cos 180
 40.0 N
o

Wk  40.0 N 20.0 m  800 J
Note that the net work is zero when the velocity is constant.
(c) If the block has a mass of 12.0 kg, what is the coefficient of kinetic
friction?
n = normal
FK
Fo
 = 36.9o
m
0  Fnet y  n  Fo sin   mg
mg = weight
n  mg  Fo sin 
Fk  k n
Fk
Fk
k 

n mg  Fo sin 
 0.457
Ex. #8: A mass of 44.0 kg is pulled towards the top of a ramp at a
constant speed by an applied force. The applied force is parallel to the
ramp, and points towards the top of the ramp. The coefficient of friction
between the ramp and the surface is 0.400 and the block is pulled a
distance of 5.00 meters up the ramp. What is the work done by each
force? What is the total work on the object? The angle of the incline is
30.0o.
n
Fo
m
Fk


mg
mg cos 
mg sin 
The first step is to solve for each force:
The ┴ forces balance:
n  mg cos 
From the definition of kinetic friction:
Fk  k n
 k mg cos 
definition
The net force parallel to the ramp is zero as well:
0  Fnet ||
  Fo  Fk  mg sin 
 Fo  k mg cos   mg sin 
The work done by the normal force is zero:
Wnormal  n x cos   n x cos 90o  0
Calculate the work done by gravity through its components.
Only the parallel component will do work.

Wgravity  mg sin  xcos 180
Wgravity  44.0kg9.80
m
o
sin 30.0 5.00m
o
s2

Wgravity  1078 J
The work done by the applied force is:
 
Wo  Fo x  cos 0
o
Fo  k mg cos   mg sin 




Fo  44.0kg9.80 m s 2  0.400cos 30.0o  sin 30.0o
Fo  365 N
Wo   365 N  5.00 m  cos  0o 
 1825 J

The work done by the friction force is:

Wdrag  k mg cos  x cos 180

o


Wdrag  0.40044.0kg9.80 m s 2 cos 30.0o 5.00m
Wdrag  747 J
Finally, the total work done is:
Wnet  1078 J  747 J  1825 J
Wnet  0
Work & Energy
Day #2: Work Energy Theorem
A 10.0- kg block of wood rests on the lab bench. A force of
50.00 N force is applied at a 30.0 degree angle above the
horizontal and it moves with constant velocity.
1. Calculate the weight.
2. Calculate the normal force.
3. How much work is done to move the block 1.00 meters?
4. Calculate the frictional work.
5. Calculate the coefficient of friction.
1. W = mg = 98N
2. N= mg- Fapplied X Sinɵ = 73 N
3. Wapplied = F‖X distance = FappliedXCosɵ
4. Wfriction = - W applied = -43 J
5. µ = Ff/N = = 43N/73N= 0.59
X 1 m = 43 J
Work Energy Theorem
I. Introduction:
A. What is energy?
Energy is defined as the ability to do work.
If an object has energy, then this object can perform
work.
Energy and work are two forms of the same overall
concept.
B. What are the types of energy?
1. kinetic energy:
Kinetic Energy is defined as energy associated with
motion. Work can be done from the movement of
an object.
wrecking
water
ball
wheel
2. potential energy:
Potential Energy is defined as energy associated
with position. Energy can be stored in an object by
virtue of its position.
pile driver
pile driver #2
C. Equation and units for kinetic energy:
Definition of kinetic energy KE:
KE  12 mv2
follow the equation for units:
KE 
1
2
massspeed 
2
m2
 kg 2
s
m
 kg  
s
2
J
work and energy are two parts of the same overall
concept, therefore they share the same units
D. What is the work energy theorem?
KE  Wnet  W
The work energy theorem states that the change of
the kinetic energy of an object is equal to the total
amount of work done on the object.
Note: the net work can be found by either
finding the sum of the work done by each
individual force or by finding the work done by
the net force.
Wnet  Fnet x cos  W  W1  W2  W3  
II. Examples:
Example #1: What is the kinetic energy of a 10.0 kg mass moving at
6.00 m/s?
KE  mv 
2
1
2
1
2
10.0 kg6.00 
m 2
s
 180 J
Ex. #2: What speed should a 2.50 kg mass have so that it has the
same kinetic energy as the above example?
2 KE
v
m
2180 J 

2.50 kg
 12.0 ms
fast solution: m and v2 are inversely proportional.
m  4
implies
v2  4 
so finally
v  2
Ex. #3: {on your own} If a third mass has a speed of 4.00 m/s, what
should its mass be so that it has the same kinetic energy as that of
problem #1?
KE  mv
1
2
2
2 KE
m 2
v
2180 J 
m
 22.5 kg
2
4.00 ms 
Ex. #4: A force of 20.0 N pushes a 10.0 kg mass for a distance of 10.0
meters. The surface is frictionless. What is the speed of the object if it
starts from rest?
Wo  Fo x cos 
Wo  20.0 N 10.0 m  cos 0o  200 J
Wnet  KE  KE  KEo
2Wnet
v
m
0
 mv  mvo
2200 J 

10.0 kg 
1
2
2
1
2
 6.32 ms
2
Ex. #5: {on your own} How much force is needed to push a 4.00 kg
mass from a speed of 5.00 m/s to a speed of 7.00 m/s in a distance of
5.00 meters?
Wnet  KE  KE  KEo
Wnet 
1
2
 mv  mvo
1
2
2
1
2
4.00 kg7.00   5.00  
m 2
s
m 2
s
Wnet  48.0 J  Fo x cos 0o
48.0 J
Fo 
 9.60 N
5.00 m
2
Ex. #6: A mass of 10.0 kg is raised by a rope with a force of 110 N.
a. What is the work done by this force if the mass is raised upwards
2.40 meters?
o
= 264 J




110
N
2
.
40
m
cos
0
Wo  Fo x cos 
b. What is the work done by gravity?
Wgravity  mgx cos 
Wgravity  10.0 kg9.80
m
2.40 mcos180
o
s2
Wgravity  235 J
c. What is the net work on the object?
Wnet  Wo  Wgravity
 264 J   235.2 J 
Wnet  28.8 J
Wnet > 0, object speeds up
d. What is the speed of the object at the end of the motion if the mass
starts from rest?
Wnet  KE  KE  KEo
2Wnet
v
m
 12 mv2  12 mvo
228.8 J 

10.0 kg
 2.40 ms
2
0
Ex. #7: A 4.00 kg mass has an initial upward velocity of 14.0 m/s. A
cord is lifting upwards on the mass. If the mass slows uniformly to a
stop in a distance of 40.0 meters, what is the lifting force?
Wnet  KE  KE  KEo
0
2
2
1
1
 2 mv  2 mvo
Wnet  392 J
Wnet  WT  Wgravity
Wnet  Tx cos 0o  mgx cos 180o
Wnet  Tx cos 0o  mgx cos 180o
Wnet  T  mg x
Wnet
T
 mg
x
 392 J

 4.00 kg 9.80 m s 2 
40.0 m
T  29.4 N
Ex. 8: A 3000 kg car has an applied force of 2500 N moving the car
forwards. At the same time, there is a 1500 N resistive force acting on
the car. a. How much distance is needed to accelerate the car from
20.0 m/s to 25.0 m/s?
Fdrag  1500 N
Fo  2500 N
Fnet   Fo  Fdrag  2500 N  1500 N
Fnet  1000 N
KE  mv  mvo
1
2

1
2
2
1
2
2
3000 kg25.0   20.0  
m 2
s
m 2
s
KE  337,500 J  337.5 kJ
KE  Wnet  Fnet x cos 0o
KE 337,500 J
 x 

 338 m
Fnet
1000 N
b. How much braking force is needed to slow the car from 25.0 m/s to a
stop in a distance of 100 meters? What is the needed coefficient of
static friction?
n
Fdrag  1500 N
Fbrake  ?
mg
KE 
1
2
3000 kg0   25.0  
m 2
s
KE  937,500 J
m 2
s
n
Fdrag  1500 N
Fbrake  ?
mg
n  mg
Fs   s n   s mg
KE  Wnet  Fdragx cos180  Fbrakex cos180
o
o
KE   Fdragx  Fbrakex
 xFdrag  Fbrake 
KE
Fbrake 
 Fdrag
 x
 937,500 J
Fbrake 
 1500 N
 100 m 
Fbrake  7875 N
Fbrake  Fs   s mg
Fbrake
s 
mg
7875 N

3000 kg9.80 m s 2 
 s  0.268
Ex. #9: If the speed of an object is doubled from speed v to speed 2v,
what happens to the kinetic energy of the object? How does the
distance compare for pushing the object from rest to speed v as
compared to pushing the object from rest to speed 2v?
KE  12 mv2  v 2
v  2
v2  4 
implies
so that
KE  4 
For accelerating an object:
1
2
mv  mvo  KE  Wnet  Fo x cos
2
1
2
2
For accelerating from 0 → v:

0
m v  vo
x 
2 Fo
2
2

mv 2

2 Fo
For accelerating from 0 → v compared to 0 → 2v:
v  2
implies
v2  4 
so that
x  4 
Ex. #10: A 1.80-kg particle has a speed of 2.0 m/s at point A and a
kinetic energy of 22.5 J at point B. What is (a) its kinetic energy at A?
(b) its speed at point B? (c) the total work done on the particle as it
moves from A to B?
KEA  mv  1.80 kg2.0
1
2
2
1
2

m 2
s
 3.60 J
2 KEB
222.5 J 
vB 

 5.00 ms
m
1.80 kg
WAB  KEB  KEA  22.5 J  3.60 J
WAB  18.9 J
Work & Energy
Day #3: Conservation of Energy
Hw: Handout #3
{back of notes packet}
Day #1: #2, 4, 6 – 10 all
Day #2: #1, 3, 5, 13
Day #3: #17 – 22 all
Potential Energy and Conservation of Energy.
I. Introduction.
A. What is potential energy?
Potential Energy is defined as energy associated
with position. Energy can be stored in an object by
virtue of its position.
B. What is the difference between a conservative and non–conservative
force?
Work is calculated by applying a force on an object
from some starting point to some finishing point.
It is possible to take different paths from a given
starting point to a given ending point.
The work done by a conservative
force is independent of the path
taken from start to finish.
A non – conservative force is not
conservative, the work does depend
on the path taken.
C. How is potential energy defined?
Potential energy is defined for conservative forces
only.
U  PE  Wc
PE = change of potential energy between two
points. {some books use U instead of PE}
Wc = work done by some conservative force
between the same two points {start to finish}
D. Derive the potential energy of gravity near Earth’s surface.
Work done in lifting an object
upwards:
Wgravity  mgx cos180
Wgravity  mgh
PE equals the minus of the work:
PE  Wgravity
  mgh
  mgh
Potential energy increases as an object rises upwards.
Work done in lowering an object
downwards:
Wgravity  mgx cos 0
Wgravity   mgh
PE equals the minus of the work:
PE  Wgravity
  mgh
  mgh
Potential energy decreases as an object lowers
downwards.
E. Derive the work done by gravity and the potential energy of gravity
on an incline.
h
sin  
x
x

mg
F  mg cos 
h
F||  mg sin 
The motion of the mass will be parallel to the ramp.
Only the parallel component of gravity will do work.
For motion to the top of the ramp:
Wgravity   F|| x
If F|| points in the opposite direction as
the displacement x, then the work is
negative.
Wgravity  mg sin  x 
Wgravity  mgh
PE  Wgravity
h 

  mg
x 
x 

This is the same result as lifting
the object straight upwards!
  mgh
  mgh
The change in the PE only depends on the vertical
height gain, not the path! conservative
In General: Potential energy increases with height.
Define the PE for any height as follows:
Let the y – axis be the vertical axis, and pick some point
to be the origin. This is the point of zero potential
energy.
The PE at any height y is given as:
PE  mgy
Since y can be negative, PE can be negative.
Example: The student has hit rock bottom (y = 0) and is
showing signs of digging.
II. Examples.
Ex. #1: A ball of mass 0.500 kg has a potential energy of 49.0 J relative
to the ground.
a. How high above the ground is the ball initially?
Let the ground level be zero height.
PE  mgy
PE
y
mg
49.0 J

0.500 kg9.80 m s 2 
y  10.0 m
b. How much kinetic energy will the ball have when it reaches the
bottom? How fast will the ball be traveling?
The ball loses 49.0 J of PE as it falls.
PE  49.0 J
PE  Wgravity  49.0 J
 Wgravity  49.0 J
0
KE  Wgravity  49.0 J  KE  KEo
 KE  49.0 J
v  14.0 ms
2KE
and v 

m
249.0 J 
0.500 kg
Ex. #2: A mass of 4.00 kg slides down an incline that measures 2.40
meters tall and 4.00 meters along the ramp. The mass starts from rest
and the ramp is frictionless. a. What is the starting potential energy of
the mass, relative to the ground?
PE only depends on vertical height. y = 0 at bottom.
PE  mgy  4.00kg 9.80 m s 2 2.40 m 
PE  94.08 J
b. What is the kinetic energy and speed of the mass at the bottom of the
ramp?
The ball loses 94.08 J of PE as it falls.
PE  94.08 J
PE  Wgravity  94.08 J
 Wgravity  94.08 J
0
KE  Wgravity  94.08 J  KE  KEo
2KE
and
 KE  94.08 J
v

m
v  6.86 ms
294.08 J 
4.00 kg
III. Force and potential energy stored in a spring.
A spring is known as a linear restoring force.
xi is the location where
the end of the spring
would sit if there were
no forces applied to
pull or push the spring.
This is the equilibrium point.
A linear force means the strength of the spring’s
push or pull is proportional to the distance x the end
of the spring is pulled from equilibrium.
The equation of the force is written in
the form:
F  kx
F = force from spring
x = distance from equilibrium
k = spring constant
The spring constant is the
relative stiffness of the
spring, measured in
newtons per meter, N/m.
A restoring force always pulls (or pushes) the end
of the spring back to the equilibrium point, to
restore the original position of the end of the
spring.
Stretching or compressing a spring stores energy
into the spring.
The amount of energy stored in a spring is:
PE  12 kx2
Omit next several slides.
IV. Conservation of Energy.
A. Derive the expression for conservation of energy for an object
moving under only conservative forces.
Start with the work energy theorem:
KE  Wnet
Separate the net work into that of conservative
forces and that of nonconservative forces.
KE  Wc  Wnc
Remember how potential energy is defined in
terms of the work done by conservative forces:
PE  Wc
Combine this with the work energy theorem:
KE  Wc  Wnc
KE  PE  Wnc
KE f  KEi  PE f  PEi   Wnc
Rearrange the terms so that everything with a
“final” is put on the left and everything with an
“initial” is put on the right:
KE f  KEi  PE f  PEi   Wnc
KE f  PE f  KEi  PEi  Wnc
If there were no nonconservative forces, then the
equation becomes:
KE f  PE f  KEi  PEi
This suggests the sum of the kinetic energies and
potential energies remains constant throughout
some process. Define the total mechanical
energy, E, of an object to be:
E  KE  PE
The energy equation becomes:
E f  Ei
This is known as the conservation of energy.
Energy is not created or destroyed during some
process, only changed from one form into
another.
B. Include nonconservative forces in the statement of conservation of
energy.
E f  Ei  Wnc
Note: For solving problems, anytime an
“applied force” is given, the work from this
force can by lumped into the work done by
nonconservative forces.
V. Examples:
Ex. #3: A spring requires a force of 40.0 N to compress it 5.00 cm. A
200 gram mass is placed against the compressed spring and released
from rest. All surfaces are frictionless.
a. What is the spring constant of the spring?
F  kx
F
k
x
40.0 N

0.0500 m
N
 800
m
b. What is the potential energy stored in the spring?
N

2



800
0
.
0500
m
PE  kx


m

1
2
2
1
2
PE  1.00 N  m  1.00 J
c. What is the speed of the projectile when it is released by the spring?
Ei  KEi  PEi
0
E f  KE f  PE f
PEi  KE f  mv
1
2
2PEi 
v
m 
21.00 J 

0.200 kg
0
2
 3.16 ms
Ex. #4: A 20.0 gram projectile is compressed against a spring with a
spring constant of 400 N/m. The spring is compressed 15.0 cm and the
projectile is launched vertically upwards. How high above its starting
position will the mass rise?
finishing height for PE of
gravity, y = ?
Spring no longer compressed.
All energy is potential energy
of gravity.
starting height for PE of
gravity, y = 0. Spring
compressed 0.150 m at this
position. All energy is
potential energy in spring.
E f  Ei  Wnc
PE f  PEi
mgy f  kxi
1
2
2

400 0.150 m
kxi
yf 

2mg
20.0200 kg 9.80 m s 2 
2
N
m
y f  23.0 m
2
Ex. #5: A pendulum is made of a heavy mass attached to and
suspended from a long, light cord. The cord for this pendulum
measures 2.00 meters long and the mass, or “bob”, measures 4.00 kg.
Take the lowest point for the mass as zero potential energy. The
pendulum is pulled to an initial angle of 30.0o from vertical and released
from rest.
a. Write an equation that gives the height of the pendulum mass from
the lowest point in terms of the length of the cord and the initial angle
from the vertical.
L  h  L cos
h  L  L cos
h  L1 cos 
b. What is the initial potential energy of the pendulum mass?
PEi  mgh  mgL1 cos 

PEi  4.00 kg9.80 m s 2 2.00 m 1  cos 30.0o
PEi  10.5 J

c. What is the speed of the mass at the bottom of the swing of the
pendulum?
starting PE turns into KE
at the bottom of the swing.
PEtop  10.5 J  KEbottom
KEbottom  mv
1
2
2
210.5 J 
v
4.00 kg
2 KEbottom
v
m
 2.29 ms
d. How far will the pendulum swing to the other side after passing the
lowest point?
Since no energy is lost, the
starting PE and the ending
PE at the extremes of the
swing are equal.
Thus the heights are equal
and the angles are equal.
Ex. #6: An Atwood’s machine has a mass, m1, of 10.0 kg suspended on
the left and a mass, m2, of 12.0 kg suspended on the right. If the
heavier mass begins 2.00 m above the ground and the lighter mass
starts at ground level, how fast will the system be traveling when the
heavier mass reaches the ground?
Initially everything starts from
rest, so no initial KE. Only m2
has PE, as it is above the ground.
m2
ytop  h
m1
ybottom  0
At the end of the motion, both
masses are moving, so both objects
have KE. Only m1 is above the
ground, so only it has PE.
Estart  m2 gh  Eend  m1v  m2 v  m1 gh
1
2
2
1
2
2
Estart  m2 gh  Eend  12 m1v 2  12 m2 v 2  m1 gh
m2 gh  m1 gh  12 m1v 2  12 m2v 2
2m2  m1 gh
v 
m2  m1 
2
212.0kg  10.0kg 9.80 m s 2 2.00m
v
12.0kg  10.0kg
v  1.89 ms
VI. Examples with Nonconservative Forces.
Ex. #7: A 2.00 kg mass is attached to a spring with a spring constant of
500 N/m. The spring is compressed 10.0 cm and the mass is released.
After passing the equilibrium position, the spring stretches to 4.00 cm
when the mass again stops. Determine the friction force and the
coefficient of kinetic friction for the mass on the support surface.
motion
Fdrag
10 cm
compression
displacement = 14 cm
4 cm
extension
equilibrium
Wnc  Fdragx cos180o
Start with energy conservation:
E f  Ei  Wnc
The mass starts and finishes at rest, so the initial
and final energy only consist of potential energy.
PE f  PEi  Wnc
1
2
Fdrag 

k x f   k xi   Fdragx
2
k xi   x f 
2
2x
2
1
2
2
  500 0.100m  0.0400m 
N
m
Fdrag  15.0 N
2
20.140m 
2
n
vertical forces balance
n  mg
Fdrag
mg
by definition:
k 
Fdrag
mg
Fdrag   k n   k mg
15.0 N

2.00 kg9.80 m s 2 
 0.765
Ex. #8: A mass of 10.0 kg starts from rest at the top of a frictionless
ramp 5.00 meters tall. At the bottom of the ramp, the mass slides on a
horizontal surface that has a coefficient of kinetic friction of 0.300.
a. What is the speed or the mass at the bottom of the ramp?
Initial energy = PE
Final energy = KE
E f  Ei  Wnc
1
2
0
mv2  KE f  PEi  mgh
v  2 gh  29.80 m s 2 5.00m 
 9.90 ms
b. How far along the surface will the mass travel before it comes to a
stop?
n
vertical forces balance
n  mg
Fdrag
motion
mg
by definition:
Fdrag   k n   k mg
Wnc  Fdragx cos180  k mgx
o
E f  Ei  Wnc
no height change along flat
ground, no change to PE
Estop  Ebottom  Wnc
KEstop  KEbottom   k mgx
0
0  mv   k mgx
1
2
x 
v
2
2 k g
2

9.90 

20.3009.80 m s 
m 2
s
2
 16.7 m
Ex. #9: A mass, m1, of 10.0 kg is set on a horizontal table with a
coefficient of kinetic friction of 0.200. A light cord ties this mass to
another mass, m2, of 6.00 kg suspended over the edge of the table. If
the suspended mass falls a distance of 1.00 meter, how fast will the pair
of masses be traveling?
m1 does not change height,
ignore its PE contribution.
m1
m2
m2 has initial height & PE,
but no height and PE at end.
Both masses start from rest and have the same
speed at the end. Use energy conservation on the
whole system.
E f  Ei  Wnc
1
2
m1 slides same
distance m2 falls
m1v  m2 v  m2 gh   k m1 gh
2
1
2
2
2m2   k m1 gh
v 
m1  m2 
2
26.00kg  0.20010.0kg 9.80 m s 2 1.00 m
v
6.00kg  10.0kg
v  2.21 ms
Power
I. Introduction.
A. Define average power and instantaneous power.
B. Equations.
Average Power is defined equal to the work done
on an object divided by the elapsed time.
Pave
W

t
Instantaneous Power is rate work is done at one
instant of time.
P  F v
W = work, t = elapsed time, F = force, v = velocity
C. Units.
Pave
W

t
PFv
J  kg m 2  1  kg m2
units    2    3
s  s  s 
s
 m   kg m  m  kg m
units  N     2    3
s
 s   s  s 
This unit is given its own special name: watt = W
m J
units  N   W  watt
s s
2
II. Examples.
Ex. #1: An automobile engine is capable of producing a maximum of
240 hp. If the car can travel at 115 mph at this full power, how much
force is the engine applying towards forward motion of the car?
Horsepower is an older unit of power, and it is
equal to 746 W. Convert values to metric:
 746 W
240 hp 
 1 hp
115
mile
hour

  179,040 W

 1609.344 m   1 hour 


 
 1mile   3600 s 
 51.41 ms
Use instantaneous power:
PFv
P
F
v
179,040 W
F
 3480 N  783 lbs
m
51.41 s
Ex. #2: An electric motor on an elevator is rated at 5.00 kW and is used
to lift 12,500 N elevator to a height of 100 meters. How much time will it
take for the elevator to rise to this height?
Pave = 5.00 kW = 5000 W
F = 12,500 N
x = 100 m
Pave
W
F x cos 0


t
t

12,500 N 100 m 
t
5000W 
F x
t
Pave
 250 s
Ex. #3: Bubba notices that his 3000 kg Lincoln slows from 80.0 mph to
60.0 mph in a time of 8.00 seconds if he lets the car coast in neutral.
What is the power needed from the motor to keep the car running at a
constant velocity of 80.0 mph?
80.0
mile
hour
 1609.344 m   1 hour 


 
 1mile   3600 s 
 35.76 ms
60.0
mile
hour
 1609.344 m   1 hour 


 
 1mile   3600 s 
 26.82 ms
Use kinematics to find the acceleration of the car:
v  vo
a
t
v  vo  at
26.82 ms  35.76 ms
a
8.00 s
 1.118 m s 2
The slowing comes from a friction force:
Fnet  ma   Fdrag
 Fdrag   ma  3000 kg 1.118 m s 2 
Fdrag  3353 N
The car must produce this same force forwards to
maintain a constant 80.0 mph.
The power output is then:
P  F v  3353 N  35.75 ms 
P  119,900W  120 kW
Converting to horsepower gives:
 1 hp 
  161 hp
P  119,900W 
 746W 
Ex. #4: The electric motor of a golf cart can accelerate the golf cart
from rest to 10.0 m/s in a time of 5.00 seconds. The mass of the cart
with passengers is 700 kg. What is the average power delivered by the
motor during the acceleration? What is the instantaneous power at the
top speed? If the motor is rated at 25 hp, what is the efficiency of the
electric motor at top speed?
Use kinematics to find the acceleration of the car
and the distance traveled:
v  vo
a
t
v  vo  at
10.0 ms  0 ms
  2.00 m s 2
a
5.00 s
x  vot  at  0 
1
2
2
1
2
2.00 5.00 s 
m
2
s2
 25.0 m
Average Power:
Pave
W
F x cos 0


t
t
Pave 
m a x

t
700 kg2.00 m s 25.0 m
2
5.00 s
Pave  7000W  7.00 kW
Instantaneous Power at Top Speed:
PFv
 mav
P  700 kg  2.00 m s 2 10.0 ms 
P  14,000W  14.0 kW
Efficiency:

Pout
14,000 W   1 hp

e

25 hp   746W
Pin



e  0.751  75.1%
Ex. #5: A fully loaded, slow moving freight elevator has a cab with a
total mass of 1,200 kg, which is required to travel upward 54 m in 3.0
minutes. The elevator’s counterweight has a mass of only 950 kg. So,
the elevator motor must help pull the cab upward. What average power
(in horsepower) is required of the force that the motor exerts on the cab
via the cable?
The counterweight drops downwards as the
elevator rises upwards, so the counterweight lifts
on the elevator with a force equal to its weight.
Flift  mcw g  mcab g
Flift  mcab g  mcw g  1200kg  950kg 9.80 m s 2 
Flift  2,450 N
The work done is:
W  Fliftx  2,450 N 54 m 
 132,300 J
The average power is:
W  132,300 J  1 min

P   
t  3.0 min  60 s

  735W

Converting units:
 1hp 
  0.985 hp
P  735W 
 746W 
Ex. #6: Mike applied 10 N of force over 3 m in 10 seconds. Joe applied
the same force over the same distance in 1 minute. Who did more
work?
Both exerted the same force over the
same displacement. Therefore, both
did the same amount of work. Time
does not matter for determining the
work done.
Ex. #7: Mike performed 5 J of work in 10 secs. Joe did 3 J of work
in 5 secs. Who produced the greater power?
Since power = work / time, we see that Mike
produced 0.5 W and Joe produced 0.6 W of power.
Thus, even though Mike did more work, he
required twice the time to do the work, and
therefore his power output was lower.
Ex. #8: Engine #1 produces twice the power of engine #2. Can we
conclude that engine #1 does twice as much work as engine #2?
No!! We cannot conclude anything about
how much work each engine does. Given
the power output, the work will depend upon
how much time is used. For example,
engine #1 may do the same amount of work
as engine #2, but in half the time.