Transcript Potential Energy Conservation of Energy

Physics 111: Mechanics
Lecture 7
Dale Gary
NJIT Physics Department
Potential Energy and
Energy Conservation








Work
Kinetic Energy
Work-Kinetic Energy Theorem
Gravitational Potential Energy
Elastic Potential Energy
Work-Energy Theorem
Conservative and
Non-conservative Forces
Conservation of Energy
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Definition of Work W

The work, W, done by a constant force on an
object is defined as the product of the component
of the force along the direction of displacement
and the magnitude of the displacement
W  ( F cos q )x



F is the magnitude of the force
Δ x is the magnitude of the
object’s displacement
q is the angle between F and x
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Work Done by Multiple Forces

If more than one force acts on an object, then
the total work is equal to the algebraic sum of
the work done by the individual forces
Wnet  Wby individual forces

Remember work is a scalar, so
this is the algebraic sum
Wnet  Wg  WN  WF  ( F cos q )r
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Kinetic Energy and Work

Kinetic energy associated with the motion of
an object
1 2
KE 
2
mv
Scalar quantity with the same unit as work
 Work is related to kinetic energy

1 2 1
mv  mv0 2  ( Fnet cos q )x
2
2
Units: N-m or J
xf
  F  dr
xi
Wnet  KEf  KEi  KE
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Work done by a Gravitational Force

Gravitational Force



Magnitude: mg
Direction: downwards to the
Earth’s center
Wnet 
1 2 1
2
mv  mv0
2
2
Work done by Gravitational
Force
W  F r cosq  F  r
Wg  mgr cos q
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Potential Energy



Potential energy is associated with the
position of the object
Gravitational Potential Energy is the
energy associated with the relative
position of an object in space near the
Earth’s surface
The gravitational potential energy
PE  mgy




m is the mass of an object
g is the acceleration of gravity
y is the vertical position of the mass
relative the surface of the Earth
SI unit: joule (J)
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Reference Levels

A location where the gravitational potential
energy is zero must be chosen for each
problem


The choice is arbitrary since the change in the
potential energy is the important quantity
Choose a convenient location for the zero
reference height
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

often the Earth’s surface
may be some other point suggested by the problem
Once the position is chosen, it must remain fixed
for the entire problem
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Work and Gravitational
Potential Energy

PE = mgy
 Wg  F y cos q  mg ( y f  yi ) cos180
 mg ( y f  yi )  PEi  PE f

Units of Potential
Energy are the same
as those of Work and
Kinetic Energy
Wgravity  KE  PE  PEi  PE f
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Extended Work-Energy Theorem

The work-energy theorem can be extended to include
potential energy:
Wnet  KEf  KEi  KE
Wgrav ity  PEi  PEf

If we only have gravitational force, then Wnet  Wgravity
KE f  KEi  PEi  PE f
KE f  PE f  PEi  KEi

The sum of the kinetic energy and the gravitational
potential energy remains constant at all time and hence
is a conserved quantity
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Extended Work-Energy Theorem

We denote the total mechanical energy by
E  KE  PE

Since

The total mechanical energy is conserved and remains
the same at all times
KE f  PE f  PEi  KEi
1 2
1 2
mvi  mgyi  mv f  mgy f
2
2
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Problem-Solving Strategy
Define the system
 Select the location of zero gravitational
potential energy



Do not change this location while solving the
problem
Identify two points the object of interest moves
between


One point should be where information is given
The other point should be where you want to find
out something
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Platform Diver
A diver of mass m drops
from a board 10.0 m above
the water’s surface. Neglect
air resistance.
 (a) Find is speed 5.0 m
above the water surface
 (b) Find his speed as he hits
the water

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Platform Diver

(a) Find his speed 5.0 m above the water
surface 1
1
mvi2  mgyi  mv2f  mgy f
2
2
1
0  gyi  v 2f  mgy f
2
v f  2 g ( yi  y f )
 2(9.8m / s 2 )(10m  5m)  9.9m / s

(b) Find his speed as he hits the water
1
0  mgyi  mv 2f  0
2
v f  2 gyi  14m / s
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Spring Force
Involves the spring constant, k
 Hooke’s Law gives the force



F  kd


F is in the opposite direction of
displacement d, always back
towards the equilibrium point.
k depends on how the spring
was formed, the material it is
made from, thickness of the
wire, etc. Unit: N/m.
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Potential Energy in a Spring

Elastic Potential Energy:



1 2
PEs  kx
2
SI unit: Joule (J)
related to the work required to
compress a spring from its
equilibrium position to some final,
arbitrary, position x
Work done by the spring
1 2 1 2
Ws   (kx)dx  kxi  kx f
xi
2
2
Ws  PEsi  PEsf
xf
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Extended Work-Energy Theorem

The work-energy theorem can be extended to include
potential energy:
Wnet  KEf  KEi  KE
Wgrav ity  PEi  PEf

Ws  PEsi  PEsf
If we include gravitational force and spring force, then
Wnet  Wgravity  Ws
( KE f  KEi )  ( PE f  PEi )  ( PEsf  PEsi )  0
KE f  PE f  PEsf  PEi  KEi  KEsi
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Extended Work-Energy Theorem

We denote the total mechanical energy by
E  KE  PE  PEs

Since

The total mechanical energy is conserved and remains
the same at all times
( KE  PE  PEs ) f  ( KE  PE  PEs ) i
1 2
1
1
1
mvi  mgyi  kxi2  mv 2f  mgy f  kx2f
2
2
2
2
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A block projected up a incline
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

A 0.5-kg block rests on a horizontal, frictionless surface.
The block is pressed back against a spring having a
constant of k = 625 N/m, compressing the spring by
10.0 cm to point A. Then the block is released.
(a) Find the maximum distance d the block travels up
the frictionless incline if θ = 30°.
(b) How fast is the block going when halfway to its
maximum height?
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A block projected up a incline
Point A (initial state): vi  0, yi  0, xi  10cm  0.1m
v f  0, y f  h  d sin q , x f  0
 Point B (final state):

d
1 2
1 2 1 2
1 2
mvi  mgyi  kxi  mv f  mgy f  kx f
2
2
2
2
1 2
kxi  mgy f  mgd sin q
2
1
2
2 kxi
mg sin q
0.5(625 N / m)( 0.1m) 2

(0.5kg)(9.8m / s 2 ) sin 30
 1.28m
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A block projected up a incline
Point A (initial state): vi  0, yi  0, xi  10cm  0.1m
v f  ?, y f  h / 2  d sin q / 2, x f  0
 Point B (final state):

1 2
1
1
1
mvi  mgyi  kxi2  mv 2f  mgy f  kx2f
2
2
2
2
1 2 1 2
h
k 2
kxi  mv f  mg ( )
xi  v 2f  gh
2
2
2
m
h  d sin q  (1.28m) sin 30  0.64m
k 2
vf 
xi  gh
m
 ......  2.5m / s
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Types of Forces

Conservative forces



Work and energy associated
with the force can be recovered
Examples: Gravity, Spring Force,
EM forces
Nonconservative forces


The forces are generally
dissipative and work done
against it cannot easily be
recovered
Examples: Kinetic friction, air
drag forces, normal forces,
tension forces, applied forces …
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Conservative Forces

A force is conservative if the work it does on an
object moving between two points is
independent of the path the objects take
between the points




The work depends only upon the initial and final
positions of the object
Any conservative force can have a potential energy
function associated with it
Wg  PEi  PE f  mgyi  mgy f
Work done by gravity
Work done by spring force
1 2 1 2
Ws  PEsi  PEsf  kxi  kx f
2
2
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Nonconservative Forces

A force is nonconservative if the work it does
on an object depends on the path taken by the
object between its final and starting points.

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The work depends upon the movement path
For a non-conservative force, potential energy can
NOT be defined
Work done by a nonconservative force
 
Wnc   F  d   f k d  Wotherforces
It is generally dissipative. The dispersal
of energy takes the form of heat or sound
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Extended Work-Energy Theorem

The work-energy theorem can be written as:
Wnet  KEf  KEi  KE
Wnet  Wnc  Wc



Wnc represents the work done by nonconservative forces
Wc represents the work done by conservative forces
Any work done by conservative forces can be accounted
for by changes in potential energy W  PE  PE
c
i
Wg  PEi  PE f  mgyi  mgy f

Gravity work

1 2 1 2
Spring force work Ws  PEi  PE f  kxi  kx f
2
2
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f
Extended Work-Energy Theorem

Any work done by conservative forces can be accounted
for by changes in potential energy
Wc  PEi  PE f  ( PE f  PEi )  PE
Wnc  KE  PE  ( KE f  KEi )  ( PE f  PEi )
Wnc  ( KE f  PE f )  ( KEi  PEi )

Mechanical energy includes kinetic and potential energy
1 2
1 2
E  KE  PE  KE  PEg  PEs  mv  mgy  kx
2
2
Wnc  E f  Ei
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Problem-Solving Strategy


Define the system to see if it includes non-conservative
forces (especially friction, drag force …)
Without non-conservative forces
1 2
1
1
1
mv f  mgy f  kx2f  mvi2  mgyi  kxi2
2
2
2
2

With non-conservative forces W  ( KE  PE )  ( KE  PE )
nc
f
f
i
i
1
1
1
1
 fd   Wotherforces  ( mv2f  mgy f  kx2f )  ( mvi2  mgyi  kxi2 )
2
2
2
2

Select the location of zero potential energy


Do not change this location while solving the problem
Identify two points the object of interest moves between


One point should be where information is given
The other point should be where you want to find out something
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Conservation of Mechanical Energy
 A block of mass m = 0.40 kg slides across a horizontal
frictionless counter with a speed of v = 0.50 m/s. It runs into
and compresses a spring of spring constant k = 750 N/m.
When the block is momentarily stopped by the spring, by
what distance d is the spring compressed?
Wnc  ( KE f  PE f )  ( KEi  PEi )
1 2
1
1
1
mv f  mgy f  kx2f  mvi2  mgyi  kxi2
2
2
2
2
1 2 1 2
0  0  kd  mv  0  0
2
2
1
1
0  0  kd 2  mv 2  0  0
2
2
d
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m 2
v  1.15cm
k
Changes in Mechanical Energy for conservative forces
A 3-kg crate slides down a ramp. The ramp is 1 m in length and
inclined at an angle of 30° as shown. The crate starts from rest at the
top. The surface friction can be negligible. Use energy methods to
determine the speed of the crate at the bottom of the ramp.
1
1
1
1
 fd   Wotherforces  ( mv2f  mgy f  kx2f )  ( mvi2  mgyi  kxi2 )
2
2
2
2
1
1
1
1
( mv 2f  mgy f  kx2f )  ( mvi2  mgyi  kxi2 )
2
2
2
2

d  1m, yi  d sin 30  0.5m, vi  0
y f  0, v f  ?
1
( mv 2f  0  0)  (0  mgyi  0)
2
v f  2 gyi  3.1m / s
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Changes in Mechanical Energy for Non-conservative forces
A 3-kg crate slides down a ramp. The ramp is 1 m in length and
inclined at an angle of 30° as shown. The crate starts from rest at the
top. The surface in contact have a coefficient of kinetic friction of 0.15.
Use energy methods to determine the speed of the crate at the bottom
of the ramp.

1
1
1
1
 fd   Wotherforces  ( mv2f  mgy f  kx2f )  ( mvi2  mgyi  kxi2 )
2
2
2
2
N
1 2
  k Nd  0  ( mv f  0  0)  (0  mgyi  0)
2
fk
 k  0.15, d  1m, yi  d sin 30  0.5m, N  ?
N  mg cos q  0
  k dmg cos q 
1 2
mv f  mgyi
2
v f  2 g ( yi  k d cos q )  2.7m / s
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Changes in Mechanical Energy for Non-conservative forces
A 3-kg crate slides down a ramp. The ramp is 1 m in length and
inclined at an angle of 30° as shown. The crate starts from rest at the
top. The surface in contact have a coefficient of kinetic friction of 0.15.
How far does the crate slide on the horizontal floor if it continues to
experience a friction force.

1
1
1
1
 fd   Wotherforces  ( mv2f  mgy f  kx2f )  ( mvi2  mgyi  kxi2 )
2
2
2
2
1
  k Nx  0  (0  0  0)  ( mvi2  0  0)
2
k  0.15, vi  2.7m / s, N  ?
N  mg  0
1
  k mgx   mvi2
2
2
v
x  i  2.5m
2 k g
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Block-Spring Collision

A block having a mass of 0.8 kg is given an initial velocity vA = 1.2
m/s to the right and collides with a spring whose mass is negligible
and whose force constant is k = 50 N/m as shown in figure. Assuming
the surface to be frictionless, calculate the maximum compression of
the spring after the collision.
1 2
1
1
1
mv f  mgy f  kx2f  mvi2  mgyi  kxi2
2
2
2
2
1 2
1 2
mvmax  0  0  mvA  0  0
2
2
xmax 
m
0.8kg
vA 
(1.2m / s)  0.15m
k
50 N / m
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Block-Spring Collision

A block having a mass of 0.8 kg is given an initial velocity vA = 1.2
m/s to the right and collides with a spring whose mass is negligible
and whose force constant is k = 50 N/m as shown in figure. Suppose
a constant force of kinetic friction acts between the block and the
surface, with µk = 0.5, what is the maximum compression xc in the
spring.
1
1
1
1
 fd   Wotherforces  ( mv2f  mgy f  kx2f )  ( mvi2  mgyi  kxi2 )
2
2
2
2
1
1
  k Nd  0  (0  0  kxc2 )  ( mvA2  0  0)
2
2
N  mg
and
d  xc
1 2 1 2
kxc  mvA    k mgxc
2
2
25 xc2  3.9 xc  0.58  0
xc  0.093m
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Energy Review
 Kinetic

Energy
Associated with movement of members of a
system
 Potential


Determined by the configuration of the system
Gravitational and Elastic
 Internal

Energy
Energy
Related to the temperature of the system
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Conservation of Energy
 Energy


is conserved
This means that energy cannot be created nor
destroyed
If the total amount of energy in a system
changes, it can only be due to the fact that
energy has crossed the boundary of the
system by some method of energy transfer
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Ways to Transfer Energy
Into or Out of A System






Work – transfers by applying a force and causing a
displacement of the point of application of the force
Mechanical Waves – allow a disturbance to propagate
through a medium
Heat – is driven by a temperature difference between
two regions in space
Matter Transfer – matter physically crosses the
boundary of the system, carrying energy with it
Electrical Transmission – transfer is by electric
current
Electromagnetic Radiation – energy is transferred by
electromagnetic waves
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Connected Blocks in Motion

Two blocks are connected by a light string that passes over a
frictionless pulley. The block of mass m1 lies on a horizontal surface
and is connected to a spring of force constant k. The system is
released from rest when the spring is unstretched. If the hanging
block of mass m2 falls a distance h before coming to rest, calculate the
coefficient of kinetic friction between the block of mass m1 and the
surface.  fd  W
 KE  PE

otherforces
1
PE  PEg  PEs  (0  m2 gh)  ( kx2  0)
2
1
  k Nx  0  m2 gh  kx2
2
N  mg
and
xh
1
  k m1 gh  m2 gh  kh2
2
1
m2 g  kh
2
k 
m1 g
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Power
Work does not depend on time interval
 The rate at which energy is transferred is
important in the design and use of practical
device
 The time rate of energy transfer is called power
 The average power is given by

W
P
t

when the method of energy transfer is work
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Instantaneous Power
Power is the time rate of energy transfer. Power
is valid for any means of energy transfer
 Other expression
W Fx

P

t

 Fv
t
A more general definition of instantaneous
power
 


W dW
dr
P  lim
t 0
t

dt
F
dt
 F v
 
P  F  v  Fv cosq
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Units of Power
 The

SI unit of power is called the watt
1 watt = 1 joule / second = 1 kg . m2 / s3
A
unit of power in the US Customary
system is horsepower

1 hp = 550 ft . lb/s = 746 W
 Units
of power can also be used to
express units of work or energy

1 kWh = (1000 W)(3600 s) = 3.6 x106 J
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Power Delivered by an Elevator Motor
 A 1000-kg elevator carries a maximum load of 800 kg. A
constant frictional force of 4000 N retards its motion upward.
What minimum power must the motor deliver to lift the fully
loaded elevator at a constant speed of 3 m/s?
Fnet , y  may
T  f  Mg  0
T  f  Mg  2.16 104 N
P  Fv  (2.16 10 4 N )(3m / s)
 6.48 10 4 W
P  64.8kW  86.9hp
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