Physics
Unit 3
 This Slideshow was developed to accompany the textbook
 OpenStax Physics
 Available for free at https://openstaxcollege.org/textbooks/college-physics
 By OpenStax College and Rice University
 2013 edition
 Some examples and diagrams are taken from the textbook.
Slides created by
Richard Wright, Andrews Academy
rwright@andrews.edu
 Which of the following is NOT work?
Pushing a Stalled Car
Pulling a Wagon
Climbing stairs
Falling Down
Carrying a Heavy Backpack Down the Hall
 Work
 The amount of force used to
move something a distance
 Want the force in the direction of
the distance
𝑊 =𝐹⋅𝑑
𝑊 = 𝐹𝑑 cos 𝜃
 Unit: 𝑁 ⋅ 𝑚 = 𝐽 (Joule)
 Watch Eureka! 08
 Marcy pulls a backpack on wheels down the 100-m hall. The 60-N
force is applied at an angle of 30° above the horizontal. How
much work is done by Marcy?
 W = 5200 J
 Mark is carrying books (200 N) down the 100-m hall. How
much work is Mark doing on the books?
W=0J
 The force is vertical
displacement is horizontal.
 You carry some books (200 N) while walking down stairs height 2
m and length 3 m. How much work do you do?
 W = -400 J
 A suitcase is hanging straight down from your hand as
you ride an escalator. Your hand exerts a force on the
suitcase, and this force does work. Which one of the
following is correct?
 The W is negative when you ride up and positive when
you ride down
 The W is positive when you ride up and negative when
you ride down
 The W is positive
 The W is negative
 Do work means  W = Fd
F = ma
 So work by a net force gives an object some acceleration
 Acceleration means the velocity changes
 𝐹 = 𝑚𝑎
 𝐹𝑑 = 𝑚𝑎𝑑
 𝑣𝑓2 = 𝑣02 + 2𝑎𝑑  solve for ad
1
 𝑎𝑑 = 2 𝑣𝑓2 − 𝑣02
1
 𝐹𝑑 = 𝑚 2 𝑣𝑓2 − 𝑣02
1
1
 𝑊 = 2 𝑚𝑣𝑓2 − 2 𝑚𝑣02
 Energy is the ability to do
work
 Kinetic Energy - Energy due to
motion
If something in motion hits
an object, it will move it
some distance
1
2
 𝐾𝐸 = 𝑚𝑣 2
Scalar
Unit is joule (J)
Watch Eureka! 09
 Work Energy Theorem
 Work of Net external force = change in kinetic energy
1
1
2
𝑊 = 𝑚𝑣𝑓 − 𝑚𝑣02
2
2
𝑊 = 𝐾𝐸𝑓 − 𝐾𝐸0
 A 0.075-kg arrow is fired horizontally. The bowstring exerts a
force on the arrow over a distance of 0.90 m. The arrow leaves the
bow at 40 m/s. What average force does the bow apply to arrow?
 Do lots of work.
 5) 3.14 × 103 𝐽
 7P1-3, 5-6, 9-13
 6) 1.30 × 103 𝐽
 Read 7.3, 7.4
 9) 9.34 × 106 𝐽, 2.33 × 109 𝐽, 250
1
 7CQ7-12
 10) 1.47
 11) 1.1 × 1010 𝐽
 Answers

1) 3 𝐽, 7.17 × 10−4

2) 1.84 × 103 𝐽
𝑚
𝑠
𝑘𝑐𝑎𝑙
 3) 5.92 × 105 𝐽, −5.88 × 105 𝐽, 0 𝐽
 12) −2470 𝑁, −1.48 × 105 𝑁
 13) −2.8 × 103 𝑁
 Potential energy
 Energy due to position
 𝑊 = 𝐹𝑑
 Gravity
 𝑊𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 𝑚𝑔ℎ
𝑃𝐸 = 𝑚𝑔ℎ
 Since the force of gravity is down
 We only worry about the vertical distance
 Potential Energy is not absolute
 It is a difference
 The path the object takes doesn’t matter, just the
vertical distance
 h is measured from any chosen point. Just be
consistent
 Watch Eureka! 10
 Conservative Forces
A force where the work it does is independent of the path
Only thing that matters is starting and stopping point
 Examples of conservative forces
 Examples of Nonconservative forces
 Gravitational Force
 Friction
 Elastic Spring Force
 Air resistance
 Electric Force
 Tension
 Normal force
 Propulsion force of things like rocket
engine
 Each of these forces depends on the path
 Potential energy can be converted
into Kinetic energy and back
 Think of an object thrown up
 Bottom  0 PE, high KE
 Top  high PE, 0 KE
 Δ𝐾𝐸 = −Δ𝑃𝐸
 𝐾𝐸𝑓 − 𝐾𝐸0 = −(𝑃𝐸𝑓 − 𝑃𝐸0 )
 Rearrange
 Conservation of Mechanical Energy
𝐾𝐸𝑓 + 𝑃𝐸𝑓 = 𝐾𝐸0 + 𝑃𝐸0
 Only if there are only
conservative forces
 If there is no work done by nonconservative forces
 Total mechanical energy is constant
KE0 + PE0 = KEf + PEf
 Other potential energy
 Spring Potential Energy
1 2
𝑃𝐸𝑆 = 𝑘𝑥
2
 A toy gun uses a spring to shoot plastic balls 𝑚 = 50 𝑔 . The spring is
compressed by 3.0 cm. Let 𝑘 = 2.22 × 105 𝑁/𝑚.
 (a) Of course, you have to do some work on the gun to arm it. How much
work do you have to do?
 (b) Suppose you fire the gun horizontally. How fast does the ball leave
the gun?
 (c) Now suppose you fire the gun straight upward. How high does the
ball go?
 You have great potential…
 18) 1.8 J, 8.6 J
 7P16-18, 20-23
 20) 0.687 m/s
 Read 7.5, 7.6
 21) 26.2 m/s, 5.35 s, 26.3 m/s, 4.86 s
 7CQ13-16
 22) 7.81 × 105 𝑁/𝑚
 23) 0.459 m
 Answers
1
 16) 1.96 × 1016 𝐽, 𝑎𝑏𝑜𝑢𝑡 2
 17) 3 × 1012 𝐽, 2 × 105 times as much
 Often both conservative and nonconservative forces act on an object at once.
 We can write Work done by net external force as
𝑊 = 𝑊𝑐 + 𝑊𝑛𝑐
𝑊𝑐 = −Δ𝑃𝐸, 𝑊 = Δ𝐾𝐸
𝑊𝑛𝑐 = Δ𝐾𝐸 + Δ𝑃𝐸
𝐾𝐸0 + 𝑃𝐸0 + 𝑊𝑛𝑐 = 𝐾𝐸𝑓 + 𝑃𝐸𝑓
𝐸0 + 𝑊𝑛𝑐 = 𝐸𝑓
 Law of Conservation of Energy
 The total energy is constant in any
process. It may change form or be
transferred from one system to
another, but the total remains the
same
 Energy is transformed from one form
to another
 Box sliding down incline
PE transformed to KE
KE transformed to Heat and
Sound
 Engine
Chemical to KE and Heat
 Efficiency
Useful energy output is always less than energy input
Some energy lost to heat, etc.
useful energy or work output 𝑊𝑜𝑢𝑡
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝐸𝑓𝑓 =
=
total energy input
𝐸𝑖𝑛
 A rocket starts on the ground at rest. Its final speed is 500 m/s
and height is 5000 m. If the mass of the rocket stays
approximately 200 kg. Find the work done by the rocket engine.
 W = 3.48 x 107 J
 A 1500-kg car’s brakes failed and it coasts down a hill from rest.
The hill is 10 m high and the car has a speed of 12 m/s at the
bottom of the hill. How much work did friction do on the car?
 Wf = -39000 J
 Captain Proton’s rocket pack provides 800,000 J of work to propel
him from resting on his ship which is near the earth to 50 m above
it. Captain Proton’s mass is 90 kg. What is his final velocity?
 V = 130 m/s
 Energy is not to be conserved while
you do this homework
 7P24-29
 Read 7.7
 7CQ18-20
 Answers
 24) 9.46 m/s
 25) 47.6 𝑚, 1.89 × 105 𝐽, 375 𝑁
 26) 4 × 104 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
 27) 24.8
𝑚
𝑠
 28) 1 × 104 𝑏𝑜𝑚𝑏𝑠
 29) 2.5 × 107 𝑦𝑒𝑎𝑟𝑠
 Two cars with the same mass do the same amount of work to get to 100
km/h.
 Which car is better
Takes 8.0 s
Takes 6.2 s
 Sometimes the time taken to do the work is important
 Rate that work is done
𝑊
𝑃=
𝑡
 Unit: joule/s = watt (W)
 Since work changes the amount of energy in an object
 Power is the rate that energy is changing
 A 500 kg car accelerates from 0 to 100 km/h in 6.2 s on a level
road. Find the average power of the car.
 P = 31100 W
41.73 horsepower
 Electrical Energy
Often measured in kWh because Pt = W
 If it costs $0.10 per kWh, how much will it cost to run a 1000 W
microwave for 2 minutes?
 Power through these problems in no time.
 33) $2.37
 7P31-40
 34) $149
 Read 7.8, 7.9
 35) 208 W, 6.57 × 109 J
 7CQ21-24
 36) 208 W, 141 s
 37) 643 hp
 Answers
 38) 3.20 s, 4.04 s
 31) 1 × 108 , 5 times, yes
 39) 7.81 × 104 W, $7.03/h, $0.00195/s
 32) 40 people, 8 million people
 40) 9.46 × 107 J, 2.54 y
 Human bodies (all living bodies)
convert energy
 Rate of food energy use is metabolic
rate
 Basal metabolic rate (BMR)
 Total energy conversion at rest
 Highest: liver and spleen
 See table 7.4
 Table 7.5 shows energy consumed
for various activites
 Energy is required to do work
 World wide, the most common source of energy is oil
 USA has 4.5% of world population, but uses 24% of world’s oil
 World energy consumption continues to increase quickly
 Growing economies in China and India
 Fossil Fuels are very polluting
 Many countries trying to develop renewable energy like wind and solar
 Generally, higher energy use per capita = better standard of living
 You have the power to change to world,
but will you work to do it?
 7P45-49, 51, 56, 63, 64
 Read 8.1, 8.2
 48) 31 g
 51) 24 W, 24.5 min, 1.14 × 106 J
 56) 144 J, 288 W
 Answers

 47) 0.100, 251 kcal
 49) 3.8 × 103 kcal
 8CQ1-5
45) 1.17 × 103
 46) 641 W, 0.860 hp
𝑊, 1.56 hp, no
 63) -21.0 N, 7.50 m, 5.77° or 84.23°, 1.50 m
 64) 1.61 × 104 N, 3.22 × 105 J, 5.66 m/s,
4.00 kJ
 Often the force acting on an object is not constant.
Baseball or Tennis ball being hit
 Times of force often short
 Force can be huge
 To hit a ball well
Both size of force and time of contact are important
Bring both these together in concept of impulse
 Impulse
 Ft
 Unit: Ns
 Is a vector
 Object responds to amount of impulse
 Large impulse  Large response  higher vf
 Large mass  less velocity
 Both mass and velocity play role in how responds to impulse
 Linear Momentum
 p = mv
 Unit: kg m/s
 Is a vector
 Is important when talking about collisions
𝐹 = 𝑚𝑎
𝑣𝑓 − 𝑣0
𝑎=
Δ𝑡
𝑣𝑓 − 𝑣0 𝑚𝑣𝑓 − 𝑚𝑣0
𝐹=𝑚
=
Δ𝑡
Δ𝑡
𝐹Δ𝑡 = 𝑚𝑣𝑓 − 𝑚𝑣0
 Impulse = Change in Momentum
 Hard to measure force during contact
 Find change in momentum
 Use impulse-Momentum Theorem and time of contact to find average force
of contact
 Watch NASCAR Crash
 A baseball (m = 0.14 kg) with initial
velocity of -40 m/s (90 mph) is hit.
It leaves the bat with a velocity of 60
m/s after 0.001 s. What is the
impulse and average net force
applied to the ball by the bat?
 Impulse = 14 Ns
 F = 14000 N
 A raindrop (m = .001 kg) hits a roof of a car at -15 m/s. After it
hits, it spatters so the effective final velocity is 0. The time of
impact is .01 s. What is the average force?
F = 1.5 N
 What if it is ice so that it bounces off at 10 m/s?
F = 2.5 N
 Watch Offset Crash
 Keep up your momentum on these
problems
 8P1-3, 7-14
 7) 9000 N
 Read 8.3
 8) 2690 N
 8CQ9-11, 13
 9) −2.40 × 103 𝑁, same
 10) 150 kg m/s, 1.43 m/s, 15 m/s
 Answers

1) 1.50 × 104
102 𝑘𝑔 ⋅ 𝑚/𝑠
𝑚
 3) 8 × 104 𝑠 , 1.20 × 106 𝑘𝑔 ⋅
𝑚
𝑚
,
−0.0100
𝑠
𝑠
𝑘𝑔 ⋅ 𝑚/𝑠, 625 times, 6.66 ×
 2) 1.20 × 108 𝑘𝑔, 1.21 × 103 𝑡𝑖𝑚𝑒𝑠
 11) -800 kg m/s, -1.20 m/s
 12) 6.67 × 103 𝑁
 13) −1.50 × 106 𝑁, −1.00 × 105 𝑁
 14) -12.0 m/s, -360 N, 0.300
 System
 All the objects involved in the problem
 Usually only two objects
 Internal Forces – Forces that the objects exert on each other
 External Forces – Forces exerted by things outside of the system
 Two balls hit in the air
 During the collision
Internal Forces = F12 and F21
External Forces = Weight (W1 and W2)
 𝐹Δ𝑡 = 𝑚𝑣𝑓 − 𝑚𝑣0
 Object 1: 𝑊1 + 𝐹12 Δt = m1 vf1 − m1 v01
 Object 2: 𝑊2 + 𝐹21 Δt = m2 vf2 − m2 v02
 Add
 𝑊1 + 𝑊2 + 𝐹12 + 𝐹21 Δt = m1 vf1 + m2 vf2 − m1 v01 + m2 v02
 𝐸𝑥𝑡 𝐹 + 𝐼𝑛𝑡 𝐹 Δ𝑡 = 𝑝𝑓 − 𝑝0
 Since F12 and F21 are equal and opposite
 Sum of internal forces = 0
 External Forces Δ𝑡 = 𝑝𝑓 − 𝑝0
 If Isolated system:
 0 = 𝑝𝑓 − 𝑝0 OR
𝑝0 = 𝑝𝑓
 Law of Conservation of Momentum
In an isolated system the total momentum remains constant
System can contain any number of objects
 Watch Crash Video
 Two billiard balls are colliding on a table. In order to apply the law of
conservation of momentum, what should the system be? One ball or both
billiard balls?
 Two billiard balls.
 External Forces: Weight and Normal Force
 If the table is horizontal these cancel
 If it were one ball, then the force of the second ball hitting it would not cancel
with anything.
 A hockey puck of mass 0.17 kg and velocity 5 m/s is caught by a .5
kg mitten laying on the ice. What is the combined velocity after
the puck is in the mitten? (ignore friction)
 v = 1.27 m/s
 A 5 kg baseball pitching machine is placed on some frictionless
ice. It shoots a 0.15 kg baseball horizontally at 35 m/s. How fast
is the pitching machine moving after it shoots the ball?
 -1.05 m/s
 This is why you feel recoil
when you shoot a gun
Reasoning Strategy
1.
Decide on the system
2.
Identify internal and external forces
3.
Is the system isolated? If no, then can’t use conservation of
momentum
4.
Set the total final momentum of the isolated system equal to the
total initial momentum




Solving problems is fun!
8P23-27
Read 8.4, 8.5
8CQ15-17
 Answers
 23) 0.122 m/s




24) 0.272 m/s
25) 2690 N, the same
26) 27.4 m/s
27) 22.4 m/s
 Watch Bumper Video
 Watch Truck Crash video
 Subatomic – kinetic energy often conserved
 Macroscopic – kinetic energy usually not conserved
 Converted into heat
 Converted into distortion or damage
 Elastic – kinetic energy conserved
 Inelastic – kinetic energy not conserved
 Completely inelastic – the objects stick together
 You are playing marbles. Your .10 kg shooter traveling at 1 m/s
hits a stationary .05 kg cat’s eye marble. If it is an elastic collision
what are the velocities after the collision?
 vc = 1.33 m/s
 vs = .333 m/s
 A ballistic pendulum can be used to determine the muzzle velocity
of a gun. A .01 kg bullet is fired into a 3 kg block of wood. The
block is attached with a thin .5m wire and swings to an angle of
40°. How fast was the bullet traveling when it left the gun?
 v = 455 m/s
 Watch Child Seat video
 Watch Reducing Risk video
 Your paper has two dimensions to
write on…use them.
 8P29-32, 34, 36, 38, 42
 Read 8.6, 8.7
 8CQ20-22
 30) -34.9 m/s, 0.150 m/s
 31) -86.4 N, -0.389 J, 64.0%
 32) 1.78 m/s, -267 J
 34) −1.05 × 10−2 m/s, 1.818 × 108 J
 36) 0.182 m/s, 8.52 × 103 J
 Answers
 38) 4.58 m/s, 31.5 J, -0.491 m/s, 3.38
J, Bigger mass = smaller KE
 29) 0.25 m/s
 42) 24.8 m/s
Collisions in 2 dimensions
 Done the same as in one dimension
 Do one equation for x and one for y
 A 2-kg cue ball is moving at 5.00 m/s and collides with a
stationary 8 ball of equal mass. After the collision the cue ball
moves at 30° to the left of the original direction and the 8-ball
moves 90° to the right of the cue balls final direction. Find the
velocities after the collision.
 vc = 4.33 m/s
 v8 = 2.5 m/s
 Rocket propulsion
Uses Newton’s 3rd Law of Motion
Forcing something out back to get an equal force forward
 As hot gasses are forced out
the back, the velocity of the
rocket increases.
 Before: 𝑝 = 𝑚𝑣
 After: 𝑚 − Δ𝑚 𝑣 + Δ𝑣 +
Δ𝑚 𝑣 − 𝑣𝑒
 Impulse: Δ𝑝 = 𝐹Δ𝑡 = −𝑚𝑔Δ𝑡
 Impulse = change in momentum
−𝑚𝑔Δ𝑡 =
𝑚 − Δ𝑚 𝑣 + Δ𝑣 + Δ𝑚 𝑣 − 𝑣𝑒
− 𝑚𝑣
−𝑚𝑔Δ𝑡 = 𝑚𝑣 + 𝑚Δ𝑣 − Δ𝑚𝑣 − Δ𝑚Δ𝑣 + Δ𝑚𝑣 − Δ𝑚𝑣𝑒 − 𝑚𝑣
−𝑚𝑔Δ𝑡 = 𝑚Δ𝑣 − Δ𝑚𝑣𝑒
Δ𝑣 Δ𝑚𝑣𝑒
−𝑔 =
−
Δ𝑡
𝑚Δ𝑡
𝑣𝑒 Δ𝑚
−𝑔 =𝑎
𝑚Δ𝑡
 Practical limit of 𝑣𝑒 = 2.5 × 103 m/s

Δ𝑚
𝑣
Δ𝑡 𝑒
= thrust
 Final velocity of one stage rocket
𝑣 = 𝑣𝑒 ln
𝑚0
𝑚𝑟
 Even though rocket accelerates, the center of mass of the rocket + gases is actually in free fall
87
 In order to leave earth 88 of the rocket must be fuel if no air resistance
 With air resistance
179
must be
180
fuel
 Try to fix using multistage rockets where the mass of the rocket decreases as the stages
are used up
 What is the acceleration of a 5000-kg rocket taking off from Mars
where the acceleration due to gravity is only 3.71 m/s2, if the
rocket expels 10 kg of gas per second at an exhaust velocity of
2.25 × 103 m/s?
 Watch curling video
 Launch into the work and rocket
through these problems.
 8P45, 47, 48, 50, 53-56
 Answers
 45) 3.00 m/s at 60° down
 47) -2.26 m/s, 7.63 × 103 J






48) 9.10 m/s at -14.7°, 0.689
50) 8.46 m/s at 33.6°, −8.78 × 104 J
53) 39.2 m/s2
54) 1.92 m/s2
55) 4.16 × 103 m/s
56) 1.60 × 104 𝑚/𝑠, 1.80 × 10−3
m/s2