Work, Energy, and
Power
Need to look at peggy’s notes
and add in more look at hers
and match them more closely.
Add in a bunch of stuff like
simple motion and some of
her early slides I like.
6.1
Work
Work is the exertion of a force in the
direction that an object moves
Work is tied to motion: No motion, no
work!
Equal to the magnitude of the force times
the magnitude of the displacement
Units are Joules (1 J = 1 N-m)
For a force in line with the motion:
W  F d
20 N
Moves
20m
Ex: A weight lifter is bench pressing a
barbell whose weight is 710N. He raises
and lowers the barbell 0.65 m at a constant
velocity. How much work is done raising
the barbell? Lowering it?
W  F  d  710 N  0.65m  461 .5 J
The work lowering the weight is -461.5J
If force is at an angle, only the portion of
the force in the direction of the motion
creates work:

F
θ
W  ( F cos )  d
Ex: How much work is done to pull a kid
in a wagon for 22 meters with a force of
40N on a handle that is at a 48 degree
angle from the ground?

W  ( F cos )  d  40N cos(48)  22m  588.835J

Force and direction of motion both
matter in defining work!


There is no work done by a force if it causes
no displacement
Forces can do positive, negative, or zero
work. When a box is pushed on a flat floor
for example
 The normal force and gravity do not
work, since they are perpendicular to the
direction of motion Cos (90º) = 0
 The person pushing the box does positive
work, since she is pushing in the direction
of motion.
Cos (0º) =+1
 Friction does negative work, since it
points opposite the direction of motion
Cos (180º) =-1

Question: If a man holds a 50 kg box at armslength for 2 hours as he
stands still, how much work does he do on the box?
FBD
W =Fcos(θ)d
Fapp
W =Fcos(θ)·0m = 0J
mg

Question: If a man holds a 50 kg box at arms length for 2 hours as he
walks 1 km forward, how much work does he do on the box?
FBD
Fapp
mg
W =Fcos(θ)d
cos(90º) = 0
W =Fcos(90º)·1000m = 0J

Question: If a man lifts a 50 kg box 2.0 meters, how
much work does he do on the box?
FBD
d = 2m
Fapp
mg
W =Fcos(θ)d
W =mgcos(θ)d
W =50kg·9.8m/s2·cos(0º)·2m = 980J








 Work Energy Theorem
Work changes mechanical energy!
Theorem relates work to this change.
Deals only with the work done by the NET
FORCE, not individual forces.
If an applied force does positive work on a
system, it tries to increase mechanical
energy.
If an applied force does negative work, it
tries to decrease mechanical energy.
The two forms of mechanical energy are
called potential and kinetic energy.
v = instantaneous velocity (initial or final) –
work is done to accelerate objects and change
their velocity


Jane uses a vine wrapped around a pulley to lift a 70-kg Tarzan to a
tree house 9.0 meters above the ground.
How much work does Jane do when she lifts Tarzan?
FBD
d = 9.0m
Wj =Fcos(θ)d
Fapp
Wj =mgcos(θ)d
mg
Wj =70kg·9.8m/s2·cos(0º)·9m = +6174J

How much work does gravity do when Jane lifts Tarzan?
Wnet = Wj+Wg = 0
Wg = -6174J


Joe pushes a 10-kg box and slides it across the floor at constant
velocity of 3.0 m/s. The coefficient of kinetic friction between the box
and floor is 0.50.
How much work does Joe do if he pushes the box for 15 meters?
15m
ΣF = 0
FBD
W =Fapp·cos(θ)d
fk
Fapp
ΣF = fk – Fapp = 0
fk = Fapp
W =fk·cos(θ)d
W =μFN·cos(θ)d
W =μmg·cos(θ)d
W =0.5·10kg·9.8m/s2·cos(0º) ·15m=735J

How much work does friction do as Joe pushes the box?
Wg = -735J
Wnet = Wj + Wg = 0

A father pulls his child in a little red wagon with constant speed. If the
father pulls with a force of 16 N for 10.0 m, and the handle of the
wagon is inclined at an angle of 60º above the horizontal, how much
work does the father do on the wagon?
ΣF = 0
θ=60º
10m
W =Fapp·cos(θ)d
W =16N·cos(60º) ·10m = 80J

Kinetic energy
 Energy due to motion
2
 K = ½ m v
 K: Kinetic Energy
 m: mass in kg
 v: speed in m/s
 Unit: Joules

How do we get to the kinetic from the equations we already
know?
2 2
Fd  12 m(a t )
KE  mv
2
1
2
Fd  m(at)
1
2
W  Fd
v  at
F  ma
Fd  mad
d  at
1
2
Fd  mv
1
2
2
W  KE f  KEo
2
Fd  ma( at )
1
2
2
2


A 10.0 g bullet has a speed of 1.2 km/s.
What is the kinetic energy of the bullet?
m=10.0g =0.010kg
v = 1.2km = 1200m
K=1/2mv2
K=1/2·0.01Kg·(1200m/s)2= 7200J

What is the bullet’s kinetic energy if the speed is halved?
K=1/2m(1/2v)2
K=1/2m1/4(v)2

K=1/4(1/2mv2)
K=1/4·7200J = 1800J
What is the bullet’s kinetic energy if the speed is doubled?
Ex: A 58 kg skier is coasting at 3.6 m/s down a 25o
slope. The slope suddenly flattens out and he slows
down to 1.2 m/s. How much work did the snow do on
the skier?
W  KE f  KE0  1 2 m v2f  1 2 m v02  0.5  58kg ((1.2 m ) 2  (3.6 m ) 2 )  334.08J
s
s
Why did the angle not impact the work done?
6.3
Work done by Gravity
Gravitational Potential Energy – energy due to an
objects position relative to the earth
The object has the potential to do work if it can fall
because of gravity
W  Fd  mad
Work gravity  PE  m  g  h
Units: Joules (J)
6.4
Conservation of Mechanical Energy
Total mechanical energy (kinetic + potential) of an
object remains constant provided the net work done is
by gravity

KEo  PEo  KE f  PE f
Ex:
One of the fastest roller coasters in the world is
the Magnum at Cedar Point. It includes a drop of
59.4 meters. Neglecting friction and starting
with a speed at the top of 0 m/s, how fast is the
coaster going at the bottom of the hill?
KEo  PEo  KE f  PE f
1/ 2mv2f  mghf  1/ 2mv02  mgh0
v f  v02  2 g (h0  h f )  (0 m s) 2  2  9.8 m s 2 (59.4m  0m)  34.121m / s
Equation from previous section:
v  v  2ax
2
2
0
Ex: A motorcycle is trying to leap across a canyon by
driving off the higher cliff at 38 m/s. Find the impact
velocity at which the cycle strikes the ground at the
top of the far side of the cliff.
38 m/s
70m
35m
1/ 2mv2f  mghf  1/ 2mv02  mgh0
v f  v02  2 g (h0  h f )  (0 m s)2  2  9.8 m s 2 (70m  35m)  26.160m / s
38m/s
θ
26.16m/s
a2 + b2 = c2
c2=√( a2 + b2)
c2=√( (38m/s)2 + (26.16m/s)2) = 46.13m/s
A=O tan θ
θ = tan-1(O/A)
θ = tan-1((26.16m/s)/(38m/s)) = 34.54º





Work Energy Theorem
Result of doing work is a change in kinetic
energy.
Theorem relates work to this change.
Deals only with the work done by the NET
FORCE, not individual forces.
The net work due to all forces equals the
change in the kinetic energy of a system.
W1+W2+W3+ …..Wn = ΔK
Wnet = ΔK

Wnet: work due to all forces acting on an
object

ΔK: change in kinetic energy (Kf – Ki)


A 15-g acorn falls from a tree and lands on the ground 10.0 m below
with a speed of 11.0 m/s.
What would the speed of the acorn have been if there had been no air
resistance and determine if 11.0m/s is real world velocity or not?
W =Fapp·cos(θ)d
v2f =v2i +2ad
vf =√((0m/s)+2 · -9.8m/s2 · 10m)= 14 m/s
W =0.015kg·9.8m/s2cos(0º) · 10m =
1.47J
KE=0.5mv2
v =√ (KE/0.5m)= √ (1.47J/(0.5·0.015kg))= 14m/s
mg d, define down direction as 0º
Did air resistance do positive, negative or zero work on the acorn?
Why?
F W =Fapp·cos(θ)d
Fapp = mg

r
W =Fr·cos(180º)d = -J
d, define down direction as 0º


A 15-g acorn falls from a tree and lands on the ground 10.0 m below
with a speed of 11.0 m/s.
How much work was done by air resistance?
Wnet = ΔK
Wg + Wd = Kf
Wnet = Wg + Wd
ΔK = Kf - Ki
Wg + Wd = Kf-Ki
Wd = 1/2mv2- Wg
Wd = 1/2·0.015kg·(11 m/s)2- 1.47J
Wd = -0.5625J
Ki = 0

What was the average force of air resistance?
Wd =Fd·cos(θ)d
Fd=Wd/(cos(θ)d)
Fd=-0.5625J / cos(180º)·10m
Fd= 0.05625 N
6.7
Power
Rate at which work is done by a force.
Rate at which energy is changing.
W F d
Power 

t
t
Net force can sometimes be found using
Newton’s 2nd law, F=ma.
If the object is moving up or down with
constant speed, the force is the weight.



SI unit for Power is the Watt.
 1 Watt = 1 Joule/s
 Named after the Scottish engineer James Watt (17761819) who perfected the steam engine.
 British system
horsepower
 1 hp = 746 W
The kilowatt-hour is a commonly used unit by the electrical
power company.
 Power companies charge you by the kilowatt-hour
(kWh), but this not power, it is really energy consumed.
 1 kW = 1000 W
 1 h = 3600 s
6
 1 kWh = 1000J/s • 3600s = 3.6 x 10 J
Ex: A 1100kg car starts from rest and accelerates for
5 seconds at 4.6m/s/s. How much power does the car
generate?
F  ma  1100kg  4.6m / s 2  5060N
d  1/ 2at2  .5  4.6m / s 2  (5s)2  57.5m
W F  d 5060 N  57.5m
Power 


 58190 J / s
t
t
5s

A record was set for stair climbing when a man ran up the
1600 steps of the Empire State Building in 10 minutes and
59 seconds. If the height gain of each step was 0.20 m, and
the man’s mass was 70.0 kg, what was his average power
output during the climb? Give your answer in both watts
and horsepower.
H of 1 step is 0.2 m
h = 0.2m · 1600 steps = 320 m
W =Fapp·cos(θ)d
Fapp =mg
W =mg·cos(θ)d
W =70kg·9.8m/s2·cos(0º) · 320m = 219520
J
t = 10min · 60s =
P= 219520J / 659s
P= W/t
600s + 59s = 659s
= 333.111 W

Calculate the power output of a 1.0 g fly as it walks straight up a
window pane at 2.5 cm/s. The window pane is 0.4 meters tall.
K=1/2mv2
m = 1.0g = 0.001kg
v = 2.5cm/s = 1 m / 100cm = 0.025m/s
K=1/2 · 0.001 kg (0.025m/s)2 = 3.13 x 10 -7 J
KE= W = 3.13 x 10-7 J
v = d/t
t=v/d
t=0.4 m / 0.025m/s = 16s
P= W/t
P= 3.13 x 10-7 J / 16s = 1.956x10-8 W
Constant force and work
The force shown is a
constant force. W =
F·Δd can be used to
calculate the work done
by this force when it
moves an object from xa
to xb. The area under the
curve from xa to xb can
also be used to calculate
the work done by the
force when it moves an
object from xa to xb
Area of a square = l·w
W = l·w =F·Δd = area
under the curve
F(x)
Δd
F
x
xa
xb
The force shown is a
variable force.
W = F·Δd CANNOT
be used to calculate the
work done by this
force!
The area under the
curve from xa to xb can
STILL be used to
calculate the work
done by the force
when it moves an
object from xa to xb
Area of a square = l·w
Area of a triangle = 0.5bh
W = l·w + 0.5bh =F·Δd =
area under the curve
F(x)
x
xa
xb



When a spring is stretched or compressed from its
equilibrium position, it does negative work, since the
spring pulls opposite the direction of motion.
 Ws = - ½ k x2
 Ws: work done by spring (J)
 k: force constant of spring (N/m)
 x: displacement from equilibrium (m)
The force doing the stretching does positive work equal to
the magnitude of the work done by the spring.
Wapp = - Ws = ½ k x2
0
F(N)
M
200
100
M
0
-100
x
-200
0
1
Fs
Ws = negative area = - ½ kx2
Fs = -kx (Hooke’s Law)
2
x(m)
3
4
5

Cont with slide 25