EnergyEnergy is anything that can be con-verted into work; i.e., anything that can exert a force through a distance.

Energy is the ability to do work.

1. Energy can be transferred from one object or system to another.

Basic Properties of Energy

2. Energy comes in multiple forms.

3. Energy can be converted from any one of these forms into any other.

4. Energy is never created anew or destroyed - this is The Law of Conservation of Energy.

Basic Types of Energy

Mechanical Energy• Energy associated with motion

Kinetic EnergyKinetic Energy: Ability to do work by virtue of motion. (Mass with velocity)A speeding car

or a space rocket

Types of Mechanical Energy

• KE = ½ mass x velocity2

• KE = ½ mv2

Examples of Kinetic Energy

2 21 12 2 (1000 kg)(14.1 m/s)K mv

2 21 12 2 (0.005 kg)(200 m/s)K mv

What is the kinetic energy of a 5-g bullet traveling at 200 m/s?

5 g

200 m/s KE = 100 JKE = 100 J

How fast must a 700 kg car drive in order to have 78,750 J of kinetic

energy?𝐾=

12𝑚𝑣2

78,750 𝐽=12(700𝑘𝑔)𝑣2

𝒗=𝟏𝟓𝒎𝒔

Potential EnergyPotential Energy: Ability to do work by virtue of position or condition.

A stretched bowA suspended weight

Types of Mechanical Energy

Gravitational Potential Energy

• PEg = weight x height• PEg = [mass X gravitational acceleration] X

height

• PEg = mgh

Elastic Potential Energy

What is the potential energy of a 50 kg person in a skyscraper if he is 480 m above the street below?

A typical 747 airplane flying at an altitude of 11 km has 2.7x1010 Joules of gravitational potential energy. What is the mass of this airplane?

Examples of Potential Energy

PE = mgh = (50 kg)(9.8 m/s2)(480 m)

PE = 235,200 JPE = 235,200 J

PE = mgh2.7x1010 J = (m)(9.8 m/s2)(11,000 m)m = 250,464 kg

Heat Energy• Energy from the internal

motion of particles of matter The hotter something is,

the faster its molecules are moving around and/or vibrating, i.e. the more energy the molecules have.

Chemical Energy• The energy from bonds between atoms or

ions

Electromagnetic Energy• Energy of moving electric charges

Nuclear Energy• Energy from the

nucleus of the atom– Fusion is when

two atoms combine• The Sun

– Fission is when the atom splits• Nuclear power

plant

Mass-Energy Equivalence: E=mc2

16

Conservation of Energy

Students will:

a) Identify situations on which conservation of mechanical energy is valid.

b) Recognize the forms that conserved energy can take.

c) Solve problems using conservation of mechanical energy.

17

Mechanical Energy

• Mechanical Energy is the sum of kinetic energy and all forms of potential energy in a system.

• In the absence of “nonconservative” resistive forces like friction and drag, mechanical energy is conserved.

• When we say that something is conserved, we mean that it remains constant.

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THE PRINCIPLE OF CONSERVATION OF MECHANICAL ENERGY

The Total Mechanical Energy (TME) of an object remains constant as the object moves, in the absence of friction.

finalinitial TMETME

ffii PEKEPEKE

constantTME

19

Con

serv

atio

n of

Ene

rgy

All Potential Energy, no Kinetic Energy

1/2 Potential Energy, 1/2 Kinetic Energy

1/4 Potential Energy, 3/4 Kinetic Energy

No Potential Energy, all Kinetic Energy

3/4 Potential Energy, 1/4 Kinetic Energy

20

If friction and wind resistance are ignored, a bobsled run illustrates how kinetic energy can be converted to potential energy, while the total mechanical energy remains constant.

21

Ski Jumping (no friction)

22

fffinal mghmvTME 221

Example 1: A person on top of a building throws a 4 kg ball upward with an initial velocity of 17 m/s from a height of 30 meters. If the ball rises and then falls all the way to the ground, what is its velocity just before it hits the ground?

17 m/s

30 m

continued on next slide

m = 4 kg vi = 17 m/s vf = ?

g = 9.8 m/s hi = 30 m hf = 0 m

iiinitial mghmvTME 221

23

ffii mghmvmghmv 2212

21

)30)(8.9)(4()17)(4( 22

21 mkgkg

sm

sm

)0)(8.9)(4()4( 22

21 mkgvkg

sm

f

JJ 1176578 Jvkg f 0)4( 221

J1754 221 )4( fvkg

2

2

877sm 2

fvsm6.29 fv

Example 1 continued:

24

Example 2: A 10 kg stone is dropped from a height of 6 meters above the ground. Find the Potential Energy, Kinetic Energy, and velocity of the stone when it is at a height of 2 meters.

6 m

At 6 m:

TME = PEi + KEi= mghi + ½ mvi2

= (10 kg)(9.8 m/s)(6 m) + ½ (10 kg)(0 m/s)2

= 588 J + 0 J

= 588 J

2 m

At 2 m:

TME = PEf + KEf = mghf + ½ mvf2

= (10 kg)(9.8 m/s)(2 m) + ½ (10 kg)(vf)2

= 196 J + ½ (10 kg)(vf)2

Therefore: 588 J = 196 J + ½ (10 kg)(vf)2

588 J – 196 J = ½ (10 kg)(vf)2

392 J = ½ (10 kg)(vf)2

solve for vf = 8.9 m/s

25

Example 3: A Daredevil MotorcyclistA motorcyclist (300 kg including the bike) is trying to leap across the canyon by driving horizontally off a cliff with an initial speed of 38.0 m/s. Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side.

38.0 m/s

70 m70 m70 m

55 m

vf = ?

26

ffii mghmvmghmv 2212

21

)70)(8.9)(300()38)(300( 22

21 mkgkg

sm

sm

)55)(8.9)(300()300( 22

21 mkgvkg

sm

f

JJ 800,205600,216 Jvkg f 700,161)300( 221

J700,260 221 )300( fvkg

2

2

1738sm 2

fvsm7.41 fv

Example 3 continued

27

•Example 4:Starting from rest, a child on a sled zooms down a frictionless slope from an initial height of 8.00 m. What is his speed at the bottom of the slope? Assume he and the sled have a total mass of 40.0 kg.

??

:

0

0

0.40

00.8

:

f

f

sm

i

i

v

Unknown

mh

v

kgm

mhh

Given8.00 m

ffii

sm

ff

fff

sm

ii

sm

ii

PEKEPEKE

JmkgmghPE

vkgmvKE

JmkgmghPE

JkgmvKE

0081.90.40

)0.40(

313900.881.90.40

000.40

2

2

2212

21

2

212

21

continued on next slide

28

fsm

fsm

fsm

f

f

f

v

v

v

vkg

J

vkgJ

vkgJJ

5.12

157

157

20

3139

203139

00.4031390

2

2

2

2 2

2

2

221

•Example 4 - continued

ffii PEKEPEKE

answer

Example 5: Unknown MassA skier starts from rest and slides down the frictionless slope as shown. What is the skier’s speed at the bottom?

H=40 m

L=250 m

start

finish

continued on next slide

m = unknown vi = 0 m/s vf = ?

g = 9.8 m/s hi = 40 m hf = 0 m

30

Example 5: Unknown Mass - continued

ffii PEKEPEKE

ffii mghmvmghmv 2212

21

You can divide the mass out of the above equation.

ffii mghmvmghmv 2212

21

)40)(8.9()0( 22

21 m

sm

sm )0)(8.9( 2

221 mv

sm

f

2

2392s

m 221

fv

sm28 fv

•Example 6:A ball is dropped from a height of 5 meters above the ground. Using conservation of energy formulas, determine the speed of the ball just before it hits the ground.