Transcript Lec 11 - Mr. Lee at Hamilton High School

Honors Physics Chapter 11
Energy and Its
Conservation
1
Chapter 11




2
Turn in Chapter 10 homework, worksheet, and
Lab Report
Take Quiz 11
Lecture
Q&A
NGSS
HS-PS3-1.
Create a computational model to calculate the
change in the energy of one component in a
system when the change in energy of the other
component(s) and energy flows in and out of the
system are known.
3
Forms of Energy

Kinetic energy: energy due to motion
–
–

Potential energy: energy stored in a system due to
state, position, or configuration
–
–
–

4
Linear kinetic energy: translational motion
Rotational kinetic energy: rotation
Gravitational potential energy: due to height of object
Spring (elastic) potential energy: due to compression or stretch
of spring
Chemical, nuclear, …
Work: energy transfer by means of force
Kinetic Energy, KE
KE 
1 2
mv
2
 m: mass
 v: velocity
Unit:
2
m2 
m
m

m
v

kg


kg

 m  N  m  Joule  J
kg


KE




 

 
2
2 
s
s 

s
2
5
Work-Kinetic Energy Theorem
Work and energy are related by
Wnet  KE  KE f  KEi
Wnet: net work done on an object
KE: kinetic energy of that object
6
Example: Pg287pp3
A comet with a mass of 7.85  1011 kg strikes Earth at a speed of
25.0 km/s. Find the kinetic energy of the comet in joules, and
compare the work that is done by Earth in stopping the comet to
the 4.2  1015 J of energy that was released by the largest nuclear
weapon ever built.
m  7.85 1011 kg , v  25.0 103
m
, Enuclear  4.2 1015 J , KE  ?
s
2
1 2 1

11
3 m
20

2.45

10
J
KE  mv    7.85  10 kg    25.0  10

2
2
s

 2.45 10 J 
20
7
Enuclear
4

5.8

10
Enuclear
15
4.2 10 J
Practice: A rifle can shoot a 4.20-g bullet at a speed of 965 m/s.
a) Find the kinetic energy of the bullet.
b) What work is done on the bullet if it starts from rest?
c) If the work is done over a distance of 0.75 m, what is the average force
on the bullet?
d) If the bullet comes to rest by pushing 1.5 cm into metal, what is the
magnitude and direction of the average force it exerts?
m
m  4.20 g  4.20 103 kg , v  965
s
a ) KE  ?
2
1 2 1
m


KE  mv   4.20 103 kg   965   1.96 103 J
2
2
s

b)W  ?
W  KE  KE f  KEi  KE f  1.96 103 J
c)d  0.75m, F  ?
W 1.96 103 J

 2.6 103 N
W  Fd  F 
d
0.75m
8
d )d  1.5cm  0.015m, KE f  0, KEi  1.96 103 J , F  ?
KEi
1.96 103 J
W  KE Fd  KE f  KEi   KEi  F  

 1.3 105 N
d
0.015m
Gravitational Potential Energy
(PE, U, or Ug)
Gravitational Potential Energy:
PE  mgh



m: mass
g = 9.8 m/s2
h = height
Unit:
m
 PE   m g h   kg  s 2   m  N  m  Joule  J
9
Frame of Reference
In the formula PE = mgh, upward has been
defined as the positive direction for h.
 We are still free to choose where h = 0.



10
Reference level, height, or point: h = 0
PE has no physical significance, only ΔPE has
physical significance, and ΔPE does not
depend on the choice of Reference level.
Example:
Is it possible for a system to have negative potential
energy?
a) No, because the kinetic energy of a system must equal
its potential energy.
b) Yes, since the choice of the zero of potential energy is
arbitrary.
c) No, because this would have no physical meaning.
d) Yes, as long as the total energy is positive.
11
Practice:
An acorn falls from a tree. Compare its kinetic
energy K, to its potential energy U.
a)
b)
c)
d)
12
K decreases and U decreases.
K increases and U decreases.
K increases and U increases.
K decreases and U increases.
Example
A 90-kg rock climber first climbs 45 m upward to the top edge
of a quarry, then, from the top, descends 85 m to the bottom.
Find the potential energy of the climber at the edge and at the
bottom, using the initial height as the reference level.
Let h = 0 at initial position.
1
h1 = 45 m
h2  45m  85m  40m,
0
h=0
PE 1  ? PE2  ?
2
h2 = ?
m  90kg , h1  45m,
PE1  mgh1  90kg  9.8m / s 2  45m  4.0 104 J
PE2  mgh2  90kg  9.8m / s 2   40m   3.5 104 J
13
Practice Pg291pp7
If a 1.8-kg brick falls to the ground from a chimney
that is 6.7 m high, what is the change in its
potential energy?
m  1.8kg , h  6.7m, PE  ?
PE  PE f  PEi  mgh f  mghi  mg  h f  hi 
m
 mg h  1.8kg  9.8 2   6.7 m   118 J
s
14
Example
A person weighing 630 N climbs up a ladder to a height of 5.0 m.
a) What work does the person do?
b) What is the increase in the gravitational potential energy of the person from
the ground to this height?
c) Where does the energy come from to cause this increase in the
gravitational potential energy?
W  Fg  630 N , h  5.0m
a ) F  Fg  630 N , d  h  5.0m,W  ?
W  Fd  630 N  5.0m  3150 J  3.2kJ
b) PE  ?
PE  mgh
 PE  mg h  630 N  5.0m  3150 J  3.2kJ
c) From person (food or fat).
15
Spring (Elastic) Potential Energy
(PEs, PEsp, or Us)
1 2
PEs  kx
2

x: compression or stretch of spring. (x is more exact.)


Displacement of spring from relaxed position
k: spring constant: Stiffness of spring , unit: N/m

Hooke’s Law: Force by spring:
Large k  stiff spring

Small k 

Unit:
16
F  kx
F is opposite to x.
soft spring
 PE    k  x  
2
N
 m 2  N  m  Joule  J
m
Example:
Calculate the work required to compress an initially
uncompressed spring with a spring constant of
25 N/m by 10 cm.
a) 0.25 J
b) 0.17 J
c) 0.10 J
d) 0.13 J
17
Practice:
What work is required to stretch a spring of spring
constant 40 N/m from x = 0.20 m to 0.25 m?
(Assume the unstretched position is at x = 0.)
a) 0.80 J
b) 0.050 J
c) 0.45 J
d) 1.3 J
18
(Total) Mechanical Energy,
Emec or E
E  KE  PE
 KE  PEg  PEsp
19
Conservation of Energy
When the only forces doing work are gravitational
forces and/or spring force during a process, the total
mechanical energy is conserved.
E E
f
 i
 KE  PE  KE  PE
i
i
f
f

 KE  PE  PE  KE  PE  PE
gi
si
f
gf
sf
 i
 E  KE  PE  0

 KE   PE
20
Example: Pg297pp15
A bike rider approaches a hill with a speed of 8.5 m/s. The combined
mass of the bike and rider is 85 kg. Choose a suitable system.
a) Find the initial kinetic energy of the system.
b) The rider coasts up the hill. Assuming there is no friction, at what
height will the bike come to rest?
c) Does your answer depend on the mass of the bike and rider?
Explain.
Let system include bike, rider, and Earth.
m
m  85kg , vi  8.5
s
a ) KEi  ?
2
1
1
m

KEi  mv 2   85kg   8.5   3070 J  3.1kJ
2
2
s

b) Let h = 0 at initial height, then hi = 0, vf = 0, hf = ?
Ei  E f  KEi  PEi  KE f  PE f
21
2
m

8.5
2


v
1
s

  3.7m
 mvi 2  mgh f  h f  i 
m
2g

2
2  9.8 2 
s 

c) No. Mass canceled out in part b).
Practice
Tarzan, mass 85 kg, swings down from a tree limb on the end
of a 20-m vine. His feet touch the ground 4.0 m below the limb.
a) How fast is Tarzan moving when he reaches the ground?
b) Does your answer depend on Tarzan’s mass?
Let h = 0 at lowest point.
m  85kg , hi  4.0m, h f  0, vi  0
l
a )v f  ?
i
Ei  E f
KEi  PEi  KE f  PE f
1
mghi  mv f 2
2
v f  2 ghi  2  9.8 m2   4.0m   8.9 m

s 
b) No, the mass canceled out.
22
s
f
h=0
Practice:
A lightweight object and a very heavy object are
sliding with equal speeds along a level frictionless
surface. They both slide up the same frictionless
hill. Which rises to a greater height?
a) The lightweight object, because it weighs less.
b) The heavy object, because it has greater kinetic
energy.
c) They both slide to the same height.
d) cannot be determined from the information given
23
Practice:
A skier, of mass 40 kg, pushes off the top of a
hill with an initial speed of 4.0 m/s.
Neglecting friction, how fast will she be
moving after dropping 10 m in elevation?
a) 7.3 m/s
b) 49 m/s
c) 15 m/s
d) 196 m/s
24
Collision

Momentum is conserved


Pi = Pf
Kinetic Energy is


Conserved  KEi = KEf  Elastic Collision
Not conserved  KEi  KEf  Inelastic Collision

vf 
25
If the two bodies stick together after collision
  v1f = v2f  Completely Inelastic Collision
m1v1i  m2 v2i
m1  m2
During completely Inelastic Collision, there is
most possible loss (but usually not all) of
kinetic energy.
Elastic Collision
Consider only collisions with v2i = 0,
Pi  Pf  P1i  P2i  P1 f  P2 f  m1v1i  m2 v2 i  m1v1 f  m2 v2 f
KEi  KE f  KE1i  KE2i  KE1 f  KE2 f

m1  m2
v1i
 v1 f 
m1  m2


 v  2m1 v
 2 f m1  m2 1i

26
2
1
1
1
1
2
2
m
v

m
v

m
v

m2 v2 f 2

1 1i
2 2i
1 1f
2
2
2
2
Only when v2i = 0
When using these formula, make sure we
choose m1 and m2 such that v2i = 0.
Example
A 2.00-g bullet, moving at 538 m/s, strikes a 0.250-kg
piece of wood at rest on a frictionless table. The
bullet sticks in the wood, and the combined mass
moves slowly down the table.
a.
Find the speed of the combination after the collision.
b.
Find the kinetic energy of the bullet before the
collision.
c.
Find the kinetic energy of the combination after the
collision.
d.
How much (percent) kinetic energy was lost?
27
Solution
m1  2.00 g  2.00  103 kg , v1i  538
a )v1 f  v2 f  v f  ?
m
, m2  0.250kg , v2i  0
s
Pi  Pf
m1v1i  m2v2i   m1  m2  v f
m

 2.00  10 kg   538 s 
m1v1i
m


4.27
vf 
m1  m2 2.00  103 kg  0.250kg
s
3
b) KE1i  ?
2
28
1
1
m

KE1i  m1v1i 2   2.00  103 kg   538   290 J
2
2
s

Solution
c ) KE f  ?
1
1
2
2
KE f  mv f   m1  m2  v f
2
2
2
1
m


  2.00  103 kg  .250kg   4.27   2.30 J
2
s

d ) KE  ?
KE % 
KE f  KEi
KEi

2.30 J  290 J
 99.2%
290
Where did the kinetic energy go?  Heat (and sound)
29
Practice
As everyone knows, bullets bounce from Superman’s chest.
Suppose Superman, mass 104 kg, while not moving, is struck by
a 4.2-g bullet moving with a speed of 835 m/s. If the collision is
elastic, find the speed that Superman had after the collision.
(Assume the bottoms of his superfeet are frictionless.)
m
m1  4.2 g  4.2  10 kg , m2  104kg , v1i  835 , v2i  0, v2 f  ?
s
3
Must choose m1 and m2 this way to use formulas.
v2 f
2   4.2  103 kg 
2m1
m
v1i 
 835

3
m1  m2
4.2  10 kg  104kg
s
m
cm
 0.0674  6.74
s
s
30
Practice: Pg300pp21
31
A 91.0-kg hockey player is skating on ice at 5.50 m/s.
Another hockey player of equal mass, moving at 8.10
m/s in the same direction, hits him from behind. They
slide off together.
a) What are the total energy and momentum in the
system before the collision?
b) What is the velocity of the two hockey players after
the collision?
c)
How much energy was lost in the collision?
m
m
m1  m2  91.0kg , v1i  5.50 , v2i  8.10 , v1 f  v2 f  v f
s
s
m
m
m1  m2  91.0kg , v1i  5.50 , v2i  8.10 , v1 f  v2 f  v f
s
s
Solution: Pg300pp21
a ) KEi  ?, Pi  ?
1
1
m1v1i 2  m2 v2i 2
2
2
2
2
1
m 1
m


  91.0kg   5.50    91.0kg   8.10   4360 J
2
s 2
s


KEi  KE1i  KE2i 
Pi  P1i  P2i  m1v1i  m2 v2i
m
m
m


  91.0kg   5.50    91.0kg   8.10   1240kg 
s
s
s


b) v f  ?
32
Pi  Pf  P1i  P2i  Pf  m1v1i  m2 v2i   m1  m2  v f
m
m
91.0kg  5.50  91.0kg  8.10
m v  m2 v2i
m
s
s
v f  1 1i

 6.8
m1  m2
91.0kg  91.0kg
s
Solution
c ) KE  ?
2
1
1
m
1

2
2
m

m
v

91.0
kg

91.0
kg
6.8




mv

KE f 
1
2
f

  4210 J
f
2
2
s

2
KE  KE f  KEi  4210 J  4360 J  150 J
33