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Chapter 10 Energy

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Chapter 10. Energy. What is Energy?. "It is important to realize that in physics today, we have no knowledge of what energy is . We do not have a picture that energy comes in little blobs of a definite amount." -Richard Feynman "Lectures on Physics". What is Energy? Con’t. - PowerPoint PPT Presentation

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Page 1: Chapter 10

Chapter 10

Energy

Page 2: Chapter 10

2AP Physics C

What is Energy?

"It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount." -Richard Feynman "Lectures on Physics"

Page 3: Chapter 10

3AP Physics C

What is Energy? Con’t

Energy - A measure of being able to do work…..–NASA.gov

Page 4: Chapter 10

4AP Physics C

Matter and Energy

The combination of matter and energy makes up the universe. Matter is substance and energy is the mover of substance.

-Paul Hewitt

Page 5: Chapter 10

5AP Physics C

Definition

Energy is an abstract quantity with the ability to effect physical change in matter.

Energy is the substance from which all things in the Universe are made up.

Page 6: Chapter 10

6AP Physics C

Forms of Energy

Two broad categories of energy:Potential: Stored Energy

Chemical – Stored in chemical bonds Mechanical – Stored in objects by tension Nuclear – Stored in nucleus of atoms Gravitational – Stored based on height and weight Electrical – Stored in batteries

Kinetic: Energy due to motion Radiant – electromagnetic Thermal - Heat Motion – Moving objects Sound – motion of air and other media by sound waves

Page 7: Chapter 10

7AP Physics C

Conservation of Energy

Energy Conservation during freefall: 2 2 2f i f iv v g y y

2 2 2 2f i f iv v gy gy

2 22 2f f i iv gy v gy

2 2 2 21 12 22 2

2 f f i i f f i im v gy v gy mv mgy mv mgy

222

2 2

ms

mmv units kg kg m Nms

m mmgy units kg m kg m Nms s

Page 8: Chapter 10

AP Physics C 8

Calculus Approach

Lets look at the freefall of a mass, m, from N2L and calculus:dvF ma m mgdt

ydv dy dv dvvdt dt dy dy

ydvmv mgdy

ymv dv mgdy

f f

i i

v y

yv ym v dv mg dy

2 2 212

1 12 2

f f

ii

v yf f i f iyv

mv mv mv mgy mgy mfy

2 21 12 2f f i imv mgy mv mgy

Page 9: Chapter 10

AP Physics C 9

Kinetic Energy

212

K mv

Page 10: Chapter 10

AP Physics C 11

Gravitational Potential Energy

Similar concept to kinetic energy except that the orientation is vertical and the force is gravity. Think KE which has not been created.

M

d

mg

Page 11: Chapter 10

AP Physics C 12

Mechanical Energy

K U k

f f i iK U K U

2 21 12 2f f i imv mgy mv mgy

K U

Page 12: Chapter 10

13AP Physics C

Problem 1

A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he releases it. Use energy to find:

a. The ball’s maximum height above the ground.b. The ball’s speed as it passes the window on its way down.c. The speed of impact on the ground.

Analysis: Use Kf + Uf = Ki + Ui to solve for y1, v2 and v3. 2 20 0 1 1

1 12 2

mv mgy mv mgy

a. At y1, v1 = 0; at y0, v0 = 10 and y0 = 0

21

1

1 10 0 02

100 5.12 9.8

m mgy

y m

Page 13: Chapter 10

AP Physics C 14

Problem 1 con’t

2 20 2

1 10 02 2

v g v g

b. At y0, v0 = 10; at y2=0, find v2

2 20 2v v

2 0 10v v

Speed up = speed down.c. At y0 = 0, v0 = 10; At y3 = -20, find v3.

2 23

1 110 0 202 2

g v g

23

12 10 20 9.8 22.22

msv

Page 14: Chapter 10

AP Physics C 15

Energy Bar Charts

Bar chart modeling a rock thrown upward and returning to same elevation:

Page 15: Chapter 10

AP Physics C 16

Zero of Potential Energy

f f

i i

U y

U yU mg dy

Our earlier calculus based derivation of The following expression represented the change in potential energy:

2 21 12 2f f i imv mgy mv mgy

The initial position was considered to be zero and resulted in U = mgyf

U will vary based on where the zero potential energy level is placed.

ΔU will always be the same regardless of the location of the zero level.

Page 16: Chapter 10

17AP Physics C

Non-Freefall Ug

Horizontal surface: no change in gravitational potential energySlanted/undulating surface: define s-axis parallel to surface of movement.

snet ss

dvF ma mdt

s s snet ss

dv dv dvdsF m m mvdt ds dt ds

sin ss

dvmg mvds

sin s smg ds mv dv

s smgdy mv dv

2 21 12 2f f i imv mgy mv mgy

Page 17: Chapter 10

18AP Physics C

Ballistic Pedulum

A 10 g bullet is fired into a 1200 g wood block hanging from a 150 cm long string. The bullet embeds itself into the block and the block swings out to an angle of 40°. What was the speed of the bullet?

Analysis: Two part problem: The impact of the bullet with the block is inelastic. Momentum is conserved. After the collision the block swings as a pendulum. The sum of the kinetic and potential energies before and after do not change as the block swings to its largest angle.

Page 18: Chapter 10

AP Physics C 19

Ballistic Pendulum con’t

1 0 0( )w B x w x B xw Bm m v m v m v

0 1W B

x xBB

m mv vm

Collision/Momentum calculations:

If we can calculate v1x from swing/energy relationships, we can calculate the speed of the bullet.

22 2

21 1

12

12

W B W B

W B W B

m m v m m gy

m m v m m gy

22 1 1

1 12 2

0 gy v gy

V2 = 0 and dividing by (mW + mB) gives:

or 1 22v gy

1 2 9.8 0.351 2.62 msv

Page 19: Chapter 10

20AP Physics C

Restoring Forces

Restoring ForceElasticEquilibrium Length, L0

Displacement, Δs Δs = L - L0

Fsp = k ΔsSpring constant, k

Page 20: Chapter 10

AP Physics C 21

Hooke’s Law

( )sp sF k s

Page 21: Chapter 10

AP Physics C 22

Hooke’s Law Problem

You need to make a spring scale for measuring mass. You want each 1.0 cm length along the scale to correspond to a mass difference of 100 g. What should be the value of the spring constant?

/ (0.100 kg)(9.8 N/m)/(0.010 m) 98 N/m k mg xsp . F k x mg

Page 22: Chapter 10

AP Physics C 23

Elastic Potential Energy

Is the force applied to the ball constant?

Describe the mechanical energy in both situations; before and after.

Page 23: Chapter 10

AP Physics C 24

Elastic Potential Energy con’t

N2L for the ball is:

net ss

dvF ma mdt

By Hooke’s law, (Fnet)s = -k(s – se), substituting gives:

edvm k s sdt

Using the chain rule:s s s

sdv dv dvds vdt ds dt ds

Substituting gives: s s emv dv k s s ds

Integrating from initial to final conditions:2 21 1 ( )

2 2s f

i i

v s

s f i ev smv ds mv mv k s s ds

Page 24: Chapter 10

AP Physics C 25

Elastic Potential Energy con’t

2 21 1 ( )2 2

s f

i i

v s

s f i ev smv ds mv mv k s s ds

From previous slide:

2 21 1( )2 2

f

i

s

e f isk s s ds k s k s

Substituting and rewriting gives:

2 22 21 1 1 12 2 2 2f f i imv k s mv k s

½ mv2 is obviously the kinetic energy. 21

2k s is the elastic potential energy, Us

Page 25: Chapter 10

AP Physics C 26

Elastic Potential Energy Problem

How far must you stretch a spring with k = 1000 N/m to store 200 J of energy?

21s 2 ( ) U k s

s2 / 2(200 J) / 1000 N/m 0.632 m s U k

Elastic potential energy is defined as:

Solving for ∆s:

Page 26: Chapter 10

AP Physics C 27

Elastic Collisions

Perfectly Elastic Collision: A collision in which mechanical energy is conserved.

Must conserve momentum and mechanical energy

Not possible where friction is involved

If only one object, m1, initially moving and all motion is along a line: fx fx ixm v m v m v 1 2 11 2 1

fx fx ixm v m v m v 2 2 2

1 2 11 2 1

1 1 12 2 2

Solving the first equation for (vfx)1 and substituting into the second gives:

ix fx fx ixmm v v m v m vm

2 1

22 22

1 2 11 21

Page 27: Chapter 10

AP Physics C 28

Elastic Collisions con’t

ix fx fx ixmm v v m v m vm

2 1

22 22

1 2 11 21

Squaring and simplifying gives:

fx fx ixmv v vm

2

2 2 11

1 2 0

There are two solutions, (vfx)2 = 0 which is trivial and:

fx ixmv v

m m

1

2 11 2

2

Substituting this into the momentum equation gives

fx ix fx ixm m mv v and v vm m m m

1 2 1

1 1 2 11 2 1 2

2

These allow us to compute the final velocity of each object in terms of the initial velocity of m1 and the relative masses of each object

Page 28: Chapter 10

AP Physics C 29

Elastic Collision Problem

A 50 g marble moving at 2.0 m/s strikes a 20 g marble at rest. What is the speed of each marble immediately after the collision?

1 2f 1 i 1

1 2

1f 2 i 1

1 2

50 g 20 g( ) ( ) (2.0 m/s) 0.86 m/s50 g 20 g

2 2(50 g)( ) ( ) (2.0 m/s) 2.9 m/s50 g 20 g

x x

x x

m mv vm m

mv vm m

Analysis: Expect that v1 will decrease and v2 will significantly increase. Laws conservation of Momentum and ME will be observed.

Page 29: Chapter 10

AP Physics C 30

Using Reference Frames

Page 30: Chapter 10

AP Physics C 31

Energy Diagrams

Energy Diagram: A graph showing a system’s potential energy and total energy as a function of position.

Page 31: Chapter 10

AP Physics C 32

Energy Diagram for a Spring

Page 32: Chapter 10

AP Physics C 33

Generalized Energy Diagram

Page 33: Chapter 10

AP Physics C 34

Stable and Unstable Equilibrium