Energy and Work - University of North Floridan00006757/physicslectures/Knight 2e/10_lect_outline/ja... · © 2010 Pearson Education, Inc. PowerPoint ... Energy and Work © 2010 Pearson

© 2010 Pearson Education, Inc.

PowerPoint® Lectures for

College Physics: A Strategic Approach, Second Edition

Chapter 10

Energy and

Work

© 2010 Pearson Education, Inc. Slide 10-2

10 Energy and Work





Reading Quiz

1. If a system is isolated, the total energy of the system

A. increases constantly.

B. decreases constantly.

C. is constant.

D. depends on work into the system.

E. depends on work out of the system.

Slide 10-6


Answer

1. If a system is isolated, the total energy of the system

A. increases constantly.

B. decreases constantly.

C. is constant.

D. depends on work into the system.

E. depends on work out of the system.

Slide 10-7


Reading Quiz

2. Which of the following is an energy transfer?

A. Kinetic energy

B. Heat

C. Potential energy

D. Chemical energy

E. Thermal energy

Slide 10-8


Answer

2. Which of the following is an energy transfer?

A. Kinetic energy

B. Heat

C. Potential energy

D. Chemical energy

E. Thermal energy

Slide 10-9


Reading Quiz

3. If you raise an object to a greater height, you are increasing

A. kinetic energy.

B. heat.

C. potential energy.

D. chemical energy.

E. thermal energy.

Slide 10-10


Answer

3. If you raise an object to a greater height, you are increasing

A. kinetic energy.

B. heat.

C. potential energy.

D. chemical energy.

E. thermal energy.

Slide 10-11


Forms of Energy

Mechanical Energy

Ug UsK

Thermal

Energy

Eth

Other forms include

Echem EnuclearSlide 10-12


The Basic Energy Model

Slide 10-13


Energy Transformations

Kinetic energy K = energy of motion

Potential energy U = energy of position

Thermal energy Eth = energy associated with

temperature

System energy E = K + U + Eth + Echem + ...

Within the System, all interactions are internal

Energy can be transformed within the system

without loss.

Energy is a property of a system.

Slide 10-14


Some Energy Transformations

Echem Ug K Eth

Echem Eth Us K Ug

Slide 10-15

Here’s another example…


Checking Understanding

A skier is moving down a slope at a constant speed. What

energy transformation is taking place?

A. K Ug

B. Ug Eth

C. Us Ug

D. Ug K

E. K Eth

Slide 10-16


A skier is moving down a slope at a constant speed. What

energy transformation is taking place?

A. K Ug

B. Ug Eth

C. Us Ug

D. Ug K

E. K Eth

Answer

Slide 10-17



A child is on a playground swing, motionless at the highest

point of his arc. As he swings back down to the lowest point of

his motion, what energy transformation is taking place?

A. K Ug

B. Ug Eth

C. Us Ug

D. Ug K

E. K Eth

Slide 10-18


Answer

A child is on a playground swing, motionless at the highest

point of his arc. As he swings back down to the lowest point of

his motion, what energy transformation is taking place?

A. K Ug

B. Ug Eth

C. Us Ug

D. Ug K

E. K Eth

Slide 10-19


Energy Transfers

These change the energy of the system

through interactions with the environment.

Work is the mechanical transfer of energy

to or from a system via pushes and pulls.

Slide 10-20

A few things to note:

•Work can be positive (work in) or negative (work out)

•We are, for now, ignoring heat…we will deal with it in Chapter 11…

•Thermal energy is…special. When energy changes to thermal energy, the

change is irreversible.


Energy Transfers: Work

W K W Eth

W Us

Slide 10-21


The Work-Energy Equation

Slide 10-22

𝐾𝑓 − 𝐾𝑖 + 𝑈𝑔 𝑓− 𝑈𝑔 𝑖

+ 𝑈𝑠 𝑓 − 𝑈𝑠 𝑖 = 𝑊

mechanical energy = 𝐾 + 𝑈𝑔 + 𝑈𝑠

Δ𝐾 + Δ𝑈𝑔 + Δ𝑈𝑠 = 𝑊

𝐾𝑓 + 𝑈𝑔 𝑓+ 𝑈𝑠 𝑓 = 𝑊 + 𝐾𝑖 + 𝑈𝑔 𝑖

+ 𝑈𝑔 𝑖


The Law of Conservation of Energy

Slide 10-23


Conservation of Mechanical Energy

Slide 10-24

𝐾𝑓 − 𝐾𝑖 + 𝑈𝑔 𝑓− 𝑈𝑔 𝑖

+ 𝑈𝑠 𝑓 − 𝑈𝑠 𝑖 = 0

Δ𝐾 + Δ𝑈𝑔 + Δ𝑈𝑠 = 0

𝐾𝑓 + 𝑈𝑔 𝑓+ 𝑈𝑠 𝑓 = 𝐾𝑖 + 𝑈𝑔 𝑖

+ 𝑈𝑠 𝑖

However… Remember that conservation of energy applies to all forms


Conceptual Example Problem

A car sits at rest at the top of a hill. A small push sends it rolling

down a hill. After its height has dropped by 5.0 m, it is moving at

a good clip. Write down the equation for conservation of energy,

noting the choice of system, the initial and final states, and what

energy transformation has taken place.

Slide 10-25


Conceptual Example Problem

A car sits at rest at the top of a hill. A small push sends it rolling

down a hill. After its height has dropped by 5.0 m, it is moving at

a good clip. Write down the equation for conservation of energy,

noting the choice of system, the initial and final states, and what

energy transformation has taken place.

Slide 10-25



Three balls are thrown off a cliff with the same speed, but in

different directions. Which ball has the greatest speed just

before it hits the ground?

A. Ball A

B. Ball B

C. Ball C

D. All balls have

the same speed

Slide 10-26


Answer




A. Ball A

B. Ball B

C. Ball C

D. All balls have

the same speed

Slide 10-27


Answer




Slide 10-27

The balls have the same speed because they start from the same height with the same 𝐾𝑖 . As each of them falls from that height, it adds an additional amount of

kinetic energy equal to the 𝑈𝑔 𝑖 (relative to the

ground) it had to begin with, so that the kinetic energy

just before it hits is 𝐾𝑓 = 𝐾𝑖 + 𝑈𝑔 𝑖. (If they had been

dropped from rest they would have gained the same additional amount of kinetic energy.) Ball C is a little different. It rises to a height higher than the top of the cliff as its 𝐾𝑖 is transformed to additional 𝑈𝑔: 𝐾𝑖 → Δ𝑈𝑔.

Then it falls, converting Δ𝑈𝑔 → 𝐾𝑖 by the time it

reaches the height of the cliff. After that it falls just like the other two balls.


Quantifying Work

Slide 10-28


Work Done by Force at an Angle to Displacement

Slide 10-29


Energy Equations

Slide 10-30

𝑊 = Δ𝐾 = 𝐾𝑓 − 𝐾𝑖

Consider the work done by wind on the sailboard:

𝑣𝑓2 = 𝑣𝑖

2 + 2𝑎𝑑

𝑊 = 𝐹𝑑 = 𝑚𝑎𝑑 ⇒ 𝑎𝑑 =𝑊

𝑚

2

𝑚𝑊 = 𝑣𝑓

2 − 𝑣𝑖2

𝑊 =1

2𝑚𝑣𝑓

2 −1

2𝑚𝑣𝑖

2


Energy Equations

Slide 10-30

It has a kinetic energy given by

Consider a point particle in a rotating object:

𝐾 =1

2𝑚𝑣2 =

1

2𝑚 𝑟𝜔 2

Sum up the 𝐾 for all the point particles in the object to get the rotational kinetic energy:

𝐾𝑟𝑜𝑡 =1

2𝑚1𝑟1

2𝜔2 +1

2𝑚2𝑟2

2𝜔2 + ⋯ =1

2∑𝑚𝑟2 𝜔2


Energy Equations

Slide 10-30

Consider an object being lifted:

Work is being done against gravity:

𝑊 = Δ𝑈𝑔 = 𝑈𝑔𝑓 − 𝑈𝑔𝑖

𝐹Δ𝑦 = 𝑈𝑔𝑓 − 𝑈𝑔𝑖

𝑚𝑔Δ𝑦 = 𝑈𝑔𝑓 − 𝑈𝑔𝑖

𝑚𝑔𝑦𝑓 − 𝑚𝑔𝑦𝑖 = 𝑈𝑔𝑓 − 𝑈𝑔𝑖


Energy Equations

Slide 10-30

Consider a spring being compressed (or stretched):

𝑊 = Δ𝑈𝑠 = 𝑈𝑠 𝑥 − 0

𝐹𝑥 = 𝑈𝑠(𝑥)

The problem here is that 𝐹 is not constant, so use average:

𝐹𝑎𝑣𝑔 =𝐹𝑖 + 𝐹𝑓

2=

0 + 𝑘𝑥

2=

1

2𝑘𝑥

𝐹𝑎𝑣𝑔𝑥 =1

2𝑘𝑥 𝑥 = 𝑈𝑠(𝑥)


Each of the boxes, with masses noted, is pulled for 10 m across

a level, frictionless floor by the noted force. Which box

experiences the largest change in kinetic energy?


Slide 10-31


Answer



experiences the largest change in kinetic energy?

D.

Slide 10-32




experiences the smallest change in kinetic energy?


Slide 10-33


Answer



experiences the smallest change in kinetic energy?

C.

Slide 10-34




experiences the largest change in speed?


Slide 10-33


Answer



experiences the largest change in speed?

C.

Slide 10-34


Example Problem

A 200 g block on a frictionless surface is pushed against a spring

with spring constant 500 N/m, compressing the spring by 2.0 cm.

When the block is released, at what speed does it shoot away

from the spring?

Slide 10-35

𝐾𝑓 = (𝑈𝑠)𝑖

1

2𝑚𝑣2 =

1

2𝑘 Δ𝑥 2 ⇒ 𝑚𝑣2 = 𝑘 Δ𝑥 2

𝑣2 =𝑘 Δ𝑥 2

𝑚 ⇒ 𝑣 =

𝑘 Δ𝑥 2

𝑚=

500 N/m 0.020 m 2

0.200 kg= 1.0 m/s


A 2.0 g desert locust can achieve a takeoff

speed of 3.6 m/s (comparable to the best

human jumpers) by using energy stored

in an internal “spring” near the knee joint.

A. When the locust jumps, what

energy transformation takes place?

B. What is the minimum amount of

energy stored in the internal spring?

C. If the locust were to make a vertical

leap, how high could it jump? Ignore

air resistance and use conservation

of energy concepts to solve this

problem.

D. If 50% of the initial kinetic energy is

transformed to thermal energy

because of air resistance, how high

will the locust jump?

Example Problem

Slide 10-36


A. The initial energy transformation is

B. There must be at least as much elastic energy in the spring as the bug’s kinetic energy moving at takeoff speed:

C. In a vertical leap the locust’s takeoff 𝐾 at the bottom would transform to 𝑈𝑔 at the top:

D. If 50% of 𝐾𝑡𝑎𝑘𝑒𝑜𝑓𝑓 is lost to air resistance:

A 2.0 g desert locust can achieve a takeoff speed of 3.6 m/s

(comparable to the best human jumpers) by using energy

stored in an internal “spring” near the knee joint.

A. When the locust jumps, what energy

transformation takes place?

B. What is the minimum amount of energy stored

in the internal spring?

C. If the locust were to make a vertical leap, how

high could it jump? Ignore air resistance and

use conservation of energy concepts to solve

this problem.

D. If 50% of the initial kinetic energy is

transformed to thermal energy because of air

resistance, how high will the locust jump?

Example Problem

Slide 10-36

𝑈𝑠 → 𝐾

𝑈𝑠 𝑚𝑖𝑛 = 𝐾𝑡𝑎𝑘𝑒𝑜𝑓𝑓

𝑈𝑠 𝑚𝑖𝑛 =1

22.0 × 10−3 kg 3.6 m/s 2

𝑈𝑠 𝑚𝑖𝑛 = 1.3 × 10−2 J

𝑈𝑔 𝑓= 𝐾𝑡𝑎𝑘𝑒𝑜𝑓𝑓

𝑚𝑔Δ𝑦 = 1.3 × 10−2 J

Δ𝑦 =1.3 × 10−2 J

2.0 × 10−3 kg 9.8 m/s2= 0.66 m

𝑈𝑔 𝑓= 0.50𝐾𝑡𝑎𝑘𝑒𝑜𝑓𝑓

𝑚𝑔Δ𝑦 = 0.65 × 10−2 J

Δ𝑦 = 0.33 m



Elastic Collisions

Slide 10-38

Using conservation of momentum and conservation of energy you get:


Power

Slide 10-39


Power

• Same mass...

• Both reach 60 mph...

Same final kinetic energy, but

different times mean different

powers.

Slide 10-40


Five toy cars accelerate from rest to their top speed in a certain

amount of time. The masses of the cars, the final speeds, and the

time to reach this speed are noted in the table. Which car has the

greatest power?

Car Mass (g) Speed (m/s) Time (s)

A 100 3 2

B 200 2 2

C 200 2 3

D 300 2 3

E 300 1 4


Slide 10-41


Answer




greatest power?


A 100 3 2

B 200 2 2

C 200 2 3

D 300 2 3

E 300 1 4

Slide 10-42





smallest power?


A 100 3 2

B 200 2 2

C 200 2 3

D 300 2 3

E 300 1 4


Slide 10-43


Answer

Four toy cars accelerate from rest to their top speed in a certain



smallest power?


A 100 3 2

B 200 2 2

C 200 2 3

D 300 2 3

E 300 1 4

Slide 10-44


In a typical tee shot, a golf ball is hit by the 300 g head of a club

moving at a speed of 40 m/s. The collision with the ball happens

so fast that the collision can be treated as the collision of a 300 g

mass with a stationary ball—the shaft of the club and the golfer

can be ignored. The 46 g ball takes off with a speed of 70 m/s.

A. What is the change in momentum of the ball?

B. What is the speed of the club head immediately after the

collision?

C. What fraction of the club’s kinetic energy is transferred to

the ball?

D. What fraction of the club’s kinetic energy is “lost” to

thermal energy?

Example Problem

Slide 10-45


In a typical tee shot, a golf ball is hit by the

300 g head of a club moving at a speed of

40 m/s. The collision with the ball happens

so fast that the collision can be treated as

the collision of a 300 g mass with a

stationary ball—the shaft of the club and

the golfer can be ignored. The 46 g ball

takes off with a speed of 70 m/s.

A. What is the change in

momentum of the ball?

B. What is the speed of the

club head immediately

after the collision?

C. What fraction of the club’s

kinetic energy is

transferred to the ball?

D. What fraction of the club’s

kinetic energy is “lost” to

thermal energy?

Example Problem

Slide 10-45

(A) Change in momentum of ball:

(B) Conservation of momentum gives:

(C)

(D) The amount of 𝐾 lost as 𝐸𝑡ℎ is the difference between the club’s initial 𝐾𝑖 and the total 𝐾𝑓 of club and ball:

Δ𝑝𝑏 = 𝑚𝑣𝑏𝑓 − 0 = 𝑚𝑣𝑏𝑓

Δ𝑝𝑏 = 0.046 kg 70m

s= 3.2

kg ⋅ m

s

𝑚𝑐𝑣𝑐𝑓 + 𝑚𝑏𝑣𝑏𝑓 = 𝑚𝑐𝑣𝑐𝑖

𝑚𝑐𝑣𝑐𝑓 = 𝑚𝑐𝑣𝑐𝑖 − 𝑚𝑏𝑣𝑏𝑓

𝑣𝑐𝑓 =𝑚𝑐𝑣𝑐𝑖 − 𝑚𝑏𝑣𝑏𝑓

𝑚𝑐

𝑣𝑐𝑓 =0.300 kg 40

ms

− 0.046 kg 70ms

0.300 kg= 29

m

s

𝐾𝑏𝑓

𝐾𝑐𝑖=

12

𝑚𝑏𝑣𝑏𝑓2

12

𝑚𝑐𝑣𝑐𝑖2

=𝑚𝑏𝑣𝑏𝑓

2

𝑚𝑐𝑣𝑐𝑖2 =

0.046 kg 70ms

2

0.300 kg 40ms

2 = 0.47

𝐾𝑖 − 𝐾𝑓

𝐾𝑖=

12

𝑚𝑐𝑣𝑐𝑖2 −

12

𝑚𝑐𝑣𝑐𝑓2 −

12

𝑚𝑏𝑣𝑏𝑓2

12

𝑚𝑐𝑣𝑐𝑖2

=240 J − 126 J − 113 J

240 J= 0.0048


A typical human head has a mass of 5.0 kg. If the head is moving

at some speed and strikes a fixed surface, it will come to rest. A

helmet can help protect against injury; the foam in the helmet

allows the head to come to rest over a longer distance, reducing

the force on the head. The foam in helmets is generally designed

to fail at a certain large force below the threshold of damage to

the head. If this force is exceeded, the foam begins to compress.

If the foam in a helmet compresses by 1.5 cm under a force of

2500 N (below the threshold for damage to the head), what is the

maximum speed the head could have on impact without

compressing the foam?

Use energy concepts to solve this problem.

Example Problem

Slide 10-46


A typical human head has a mass of

5.0 kg. If the head is moving at some

speed and strikes a fixed surface, it

will come to rest. A helmet can help

protect against injury; the foam in the

helmet allows the head to come to

rest over a longer distance, reducing

the force on the head. The foam in

helmets is generally designed to fail

at a certain large force below the

threshold of damage to the head. If

this force is exceeded, the foam

begins to compress.

If the foam in a helmet compresses

by 1.5 cm under a force of 2500 N

(below the threshold for damage to

the head), what is the maximum

speed the head could have on impact

without compressing the foam?

Use energy concepts to solve this

problem.

Example Problem

Slide 10-46


Data for one stage of the 2004 Tour de France show that Lance

Armstrong’s average speed was 15 m/s, and that keeping Lance

and his bike moving at this zippy pace required a power of 450 W.

A. What was the average forward force keeping Lance and

his bike moving forward?

B. To put this in perspective, compute what mass would

have this weight.

Example Problem

Slide 10-47


Summary

Slide 10-48


Summary

Slide 10-49


Trucks with the noted masses moving at the noted speeds crash

into barriers that bring them to rest with a constant force. Which

truck compresses the barrier by the largest distance?

Additional Questions

Slide 10-54




truck compresses the barrier by the largest distance?

Answer

E.

Slide 10-55




truck compresses the barrier by the smallest distance?


Slide 10-56




truck compresses the barrier by the smallest distance?

Answer

B.

Slide 10-57


A 20-cm-long spring is attached to a wall. When pulled

horizontally with a force of 100 N, the spring stretches to a

length of 22 cm. What is the value of the spring constant?

A. 5000 N/m

B. 500 N/m

C. 454 N/m


Slide 10-58


Answer

A 20-cm-long spring is attached to a wall. When pulled

horizontally with a force of 100 N, the spring stretches to a

length of 22 cm. What is the value of the spring constant?

A. 5000 N/m

B. 500 N/m

C. 454 N/m

Slide 10-59


I swing a ball around my head at constant speed in a circle with

circumference 3 m. What is the work done on the ball by the 10 N

tension force in the string during one revolution of the ball?

A. 30 J

B. 20 J

C. 10 J

D. 0 J


Slide 10-60


Answer

I swing a ball around my head at constant speed in a circle with

circumference 3 m. What is the work done on the ball by the 10 N

tension force in the string during one revolution of the ball?

A. 30 J

B. 20 J

C. 10 J

D. 0 J

Slide 10-61

Documents

Energy and Work - University of North Floridan00006757/physicslectures/Knight 2e/10_lect_outline/ja... · © 2010 Pearson Education, Inc. PowerPoint ... Energy and Work © 2010 Pearson