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WORK AND ENERGY
Scalars are back
REVIEW

Equations for Motion Along One Dimension
x
vave 
t
x dx
v  lim

t  0 t
dt
v
aave 
t
v dv
a  lim

t 0 t
dt
REVIEW

Motion Equations for Constant Acceleration
• 1.
v  v0  at
• 2. x  x0  v0t 
1
• 3. v  v  2ax
2
• 4. vave
2
0
v  v0

2
2
at
2
REVIEW
3 Laws of Motion
 If in Equilibrium

F  0
If not in equilibrium
 Change in Motion is Due to Force

 F  ma

Force causes a change in acceleration
SPRINGS AND OTHER PROBLEMS
Force exerted by a
spring is dependent on
amount of deformity
of the spring
 Amount of force
applied changes
continuously over
time
 What is the velocity of
an object launched
from the spring?

WORK
Work done on an
object by all forces is
equal to the change in
kinetic energy of the
object.
 This definition is valid
even if the force is not
constant

WORK – CONSTANT FORCE
When a force, F, is
doing work on an
object, the object will
move and be
displaced.
 The work done, by the
force, F, is defined as

W  Fd

Where d is the objects
displacement
WORK – CONSTANT FORCE

We are only interested
in the component of
the force that is
parallel to the
direction of motion
W  F||d
WORK – CONSTANT FORCE

We are only interested
in the component of
the force that is
parallel to the
direction of motion
W  Fd cos

or
 
W  F d
JOULE
W  Fd cos

Work done by 1N of
force to move an object
1 meter in the same
direction
W  1N 1m
W  1N  m
W  1Joule  1J
JAMES PRESCOTT JOULE
December 24, 1818October 11, 1889
 The mechanical
equivalent of heat


838 ft.lbf of work to
raise temperature of 1
lb of water by 1 degree
farenheit
Led to the theory of
conservation of energy
 Helped Lord Kelvin
develop the absolute
scale of temperature

WORK – ZERO, NEGATIVE, POSITIVE

When defining work
done, its always
important to specify
which force is acting
on what object



Work done by man
Work done by gravity
Work done by barbell
TOTAL WORK
Compute work done
by forces individually
 Then just add to get
total work done on the
object

Wtot  W  W1  W2  ...

Note: work is scalar
EXAMPLE

Farmer hitches a tractor
with firewood and pulls
it a distance 20m on
level ground. Total
weight of the sled and
wood is 14700N and the
tractor pulls with a
constant force of 5000N
at an angle 36.9o above
the horizontal. There is
a 3500N friction force
opposing the motion.
Find the work done by
each of the forces and
the total work done by
all forces.
EXAMPLE
Wtractor  FTractord cos 
Wtractor  (5000)( 20) cos(36.9)
Wtractor  79968
Wtractor  80000 J
W friction  F frictiond cos 
W friction  (3500)( 20) cos(180)
W friction  70000 J
WORK DONE BY NON-CONSTANT FORCE

Requires the use of integrals
ENERGY
Energy is a hard to
define concept
 Simplified definition
 The ability of a
physical system to do
work on another
physical system
 Many types of energythese are much easier
to define

KINETIC ENERGY
Energy of motion
 When work is done to
an object the object
moves
 It also affects an
objects speed

W>0 – object speeds
up
 W<0 – object slows
down
 W=0 – no effect

KINETIC ENERGY
W  Fd

Newton’s 2nd Law
W  mad
KINETIC ENERGY
W  mad
v 2  v02  2ad
v 2  v02  2ad
v 2  v02
a
2d
 v 2  v02 
d
W  m
 2d 
 v 2  v02 

W  m
 2 
1 2 1 2
W  mv  mv0
2
2
KINETIC ENERGY
1 2 1 2
W  mv  mv0
2
2

Work done is the
change in kinetic
energy of an object
1 2
K  mv
2
 This is translational
kinetic energy
WORK – ENERGY THEOREM

Assuming mass is
constant
1 2
W   mv
2
W  K



Unit of work is Joules
Unit of energy is also
Joules
Note: Energy is also
scalar
EXAMPLE

Farmer hitches a tractor
with firewood and pulls
it a distance 20m on
level ground. Total
weight of the sled and
wood is 14700N and the
tractor pulls with a
constant force of 5000N
at an angle 36.9o above
the horizontal. There is
a 3500N friction force
opposing the motion.
Suppose it’s initial
speed is 2.0 m/s, what is
its final speed after
travelling 20m.
EXAMPLE
Wtotal  10000 J
Wtotal  KE
1
2
2
Wtotal  m(v  v0 )
2
1
2
2
10000  (1500)(v  2 )
2
13.333  v 2  2 2
v  4.1633
v  4ms
EXAMPLE

A 15kg block is placed
on a 40o incline and
allowed to slide for
5m. What is it’s final
speed?
15kg
POTENTIAL ENERGY
Energy due to a body’s
configuration or
surroundings.
 Many different types

Springs
 Electrical
 Gravitational

GRAVITATIONAL POTENTIAL
An object held in the
air has the “potential”
to do work once
released.
 Assume object at some
height
v0  0


After travelling some
distance
y2
2
v  v0  2ad
v 2  2(  g )(  y )
v 2  2 gy
GRAVITATIONAL POTENTIAL
An object held in the
air has the “potential”
to do work once
released.
 KE after travelling
some distance y

v 2  2 gy
1 2 1
K  mv  2mgy
2
2
K  mgy
GRAVITATIONAL POTENTIAL
An object held in the
air has the “potential”
to do work once
released.
 Amount of potential
work

K0  0
K  mgy
W  K  mgy
GRAVITATIONAL POTENTIAL

An object held in the
air has the “potential”
to do work once
released.
PEgrav  U  mgy

Note: choose your
origin and be
consistent
EXAMPLE- GIANCOLI 6-28

By how much does the gravitational potential
energy of a 64-kg pole vaulter change if his center
of mass rises 4.0m?
EXAMPLE- GIANCOLI 6-28

By how much does the gravitational potential
energy of a 64-kg pole vaulter change if his center
of mass rises 4.0m?
U  mgy
U 0  mg (0)  0
U  mg (4)  (64)(9.8)( 4)
U  2500 J
U  2500 J
WORK DONE EXAMPLE
What is work done to
lift a block by 5 m?
 If a 40o was used?

15kg
CONSERVATIVE AND NON-CONSERVATIVE
FORCE
Conservative Force
 Work Done is
independent of the
path taken

Non Conservative
Force
 Work done depends on
the path taken

Gravity
 Elastic
 Electric
Friction
 Air resistance
 Tension
 Push-Pull from a
person


You can “store” energy
in these types of
systems by doing work
on the system


Cannot define
potential energy for
these types of forces
CONSERVATION OF MECHANICAL ENERGY

If only gravity is
acting on the object
W  K
W  Wgrav   U
K    U
K  K 0  (U  U 0 )
K  U  K0  U 0

Valid for all
conservative forces

If only conservative
forces are acting, the
total mechanical
energy of a system
neither increase nor
decrease in any
process. It stays
constant- it is
conserved.
CONSERVATION OF MECHANICAL ENERGY

If a non-conservative
force is acting on the
object
W  Wgrav  WNC
W  Wgrav  WNC  K
Wgrav  U
K  WNC  U
WNC  K  U
K  U  K 0  U 0  WNC

Most common nonconservative energy is
friction
WNC   fd
EXAMPLE – FROM OUR 2ND LECTURE

A motorcycle stuntman rides over a cliff. Just at
the cliff edge his velocity is completely horizontal
with magnitude 9.0 m/s. Find the motorcycles
speed after 0.50s.
LIST THE GIVEN
Origin is cliff edge
 a=-g=-9.80m/s2
 At time t=0s



y0  0
x0  0

v0  9.0 m s

At time t=0.50s
d  ?
v ?

v0

v
SPLIT INTO COMPONENTS

v0 x  9.0 m s
v0 y  0
 

D  Dx  D y
Dx  x
Dy  y
  
v  vx  v y

v0

v
CALCULATE COMPONENTS
INDEPENDENTLY
vx  v0 x  9.0 m s
v y  v0 y  gt

v0

vx
v y   gt  (9.8)(0.5)
v y  4.9 m s

vy

v
CALCULATE VELOCITY
vx  9.0 m s
v y  4.9 m s
2
2
v  vx  v y
v  (9.0)  (4.9)
2

v0

vx
2
v  10.25 m s  1.0 x10 m s

vy

v
NOT NEEDED
vx  9.0 m s
v y  4.9 m s
v  1.0 x10 m s
v y  4.9
tan   
 0.544
vx
9
  28.56  29
v  1.0 x10 m s

29o below the horizontal

v0

vy


vx

v
ALTERNATE SOLUTION
K  U  K0  U 0
v0  9.0 m s
K 0  1 2 mv02

v0
U0  0

vx

vy

v
ALTERNATE SOLUTION
K  U  K0  U 0
v0  9.0 m s

v0
K 0  1 2 mv02
U0  0
y  y0  v0 y t 
y
1
1
2
gt
2
gt   2 (9.8)(0.5)
2
2

vx
y  1.225m
1
2

vy

v
ALTERNATE SOLUTION
K  U  K0  U 0
K0 
1
2
mv
2
0
U0  0
U  mgy

v0

vx
K  K0  U 0  U
1
2
mv

2
1
2
mv
2
0  0  mgy
v  v  gy
2
2
0
v 2  9 2  2(9.8)( 1.225)
v 2  105.1
v  10.247

vy

v
PROBLEM – YOUNG AND FREEDMAN 7.14

A small rock with mass 0.12 kg is fastened to a
massless string with length 0.80 m to form a
pendulum. The pendulum is swung so that it
makes a maximum angle of 45o with the vertical.
(a) What is the speed of the rock when it passes
the vertical position? (b) What is the tension in
the string when it makes an angle 45o with the
vertical? (c) What is the tension in the string
when it passes through the vertical?
PROBLEM – SERWAY 7.33

A crate of mass 10.0 kg is pulled up a rough
incline with an initial speed of 1.50 m/s. The
pulling force is 100N parallel to the incline,
which makes an angle of 20o with the horizontal.
The coefficient of kinetic friction is 0.400, and the
crate is pulled 5.00m. (a) How much work is done
by the gravitational force on the crate? (b)
Determine the increase in internal energy of the
crate-incline system due to friction. (c) How much
work is done by the 100N force on the crate? (d)
What is the change in kinetic energy of the crate?
(e) What is the speed of the crate after being
pulled 5m?
OTHER TYPES OF POTENTIAL ENERGY
Elastic Potential
 For Ideal Springs
 If a spring is to be
stretched a certain
distance x

F  kx

Where k is the spring
constant (the spring’s
stiffness)

It’s me again
POTENTIAL ENERGY OF SPRINGS
Restoring Force
Fs  kx
 Hooke’s Law – valid
for small x

POTENTIAL ENERGY OF SPRINGS

Work done ON the
spring (from
equilibrium)
F  kx
W  Fd
W  (kx)( x)




NO
Force is not constant
We can still use average
force
Luckily F varies linearly
with x
POTENTIAL ENERGY OF SPRINGS

Work done ON the spring
(from equilibrium)
F0  k (0)  0
F  kx
Fave 
1
2
( F  F0 )
W  Fave d
W  ( 1 2 kx)( x)

W
1
U
1
2
kx
2
2
kx
2
Where U is the elastic
potential
CONSERVATION OF MECHANICAL ENERGY
EXPANDED

Conservative
K  U grav  U spring  K 0  U 0 grav  U 0 spring
With Non conservative
K  U grav  U spring  K 0  U 0 grav  U 0 spring  WNC

YOUNG AND FREEDMAN 7.20

A 1.20kg piece of
cheese was placed on
a vertical spring of
negligible mass and
force constant k=1800
N/m that is
compressed 15.0 cm.
When the spring is
released how high
does the cheese rise
from its original
position?
POWER

Rate at which work is
done
PAve
Work

Time
SI unit is called the
Watt = 1J/s
 Horsepower =
550ftlb/s = 746W

POWER

Rate at which work is
done
Work
PAve 
Time
Fd
PAve 
 Fvave
t

Efficiency
P
e  out
Pin
EXAMPLE GIANCOLI 6-58

How long will it take a 1750W motor to lift a 315
kg piano to a sixth story window 16.0m above?
EXAMPLE GIANCOLI 6-58

How long will it take a 1750W motor to lift a 315
kg piano to a sixth story window 16.0m above?
Work
PAve 
 1750
Time
W  Fd  315(9.8) 16  49392 J
W
t
 28.2s
P
PROBLEM SERWAY 7.40

A 650 kg elevator starts from rest. It moves
upward for 3s with constant acceleration until it
reaches its cruising speed of 1.75m/s. (a) What is
the average power of the elevator motor during
this period? (b) How does this compare when the
elevator moves at cruising speed?
YOUNG AND FREEDMAN – 7.42

A 2.00 kg block is pushed against a spring with
negligible mass and force constant k= 400 N/m,
compressing it 0.220m. When the block is
released, it moves along a frictionless, horizontal
surface and then up a frictionless incline with
slope 37.0o. (a) what is the speed of the block on
the horizontal surface after leaving the spring?
(b) How far up the slope does the block travel
before starting to slide back down?
GIANCOLI 6-56

A 280 g wood block is firmly attached to the end
of a horizontal spring. The block can slide along
the table with a coefficient of friction of 0.30. A
force of 22 N compresses the string 18 cm. if the
spring is released, how far from the equilibrium
position will it stretch at its first maximum
extension.
GROUP WORK

A 1500 kg rocket is to be launched with an initial
upward speed of 50.0 m/s. In order to assist the
engines, the engineers will start it from rest on a
ramp that rises 53o above the horizontal. The
engines provide a constant forward thrust of
2000N and the coefficient of kinetic friction with
the ramp is 0.05. At what height should the
rocket start?