Types of Energy How many types/forms of energy can How many types/forms of energy can

you list?you list?

Gravitational Potential Energy (PE)

The work done to lift (or lower) something is equal to its change in gravitational potential energy

PE = Fd = (mg)Δh

= mg (h2 – h1)

PE=mgh

PE: in Joules

m: in kg

h: in m

h h

mm

h=0 (reference point)

Example:Find the potential energy of a 60.0 kg student sitting on a 80.0 cm high stool…

a) relative to the floor in the room

b) relative to the floor in the room 1 floor below (3.00 m lower)

c) relative to the ceiling 2.50 m above the student

Example:Find the potential energy of a 60.0 kg student sitting on a 80.0 cm high stool…

a) relative to the floor in the room.

b) relative to the floor in the room 1 floor below (3.0 m lower)

c) relative to the ceiling 2.50 m above the student

a) PE = mgh

= 60.0 x 9.8 x 0.800

= 470 Joules

b) PE = mgh

= 60.0 x 9.8 x 3.8

= 2200 Joules

c) PE = mgh

= 60 x 9.8 x (-2.5)

= -1470 Joules

Kinetic Energy (KE) The energy of motion. Moving things

have kinetic energy. Proportional to mass of the object and the square of its velocity.

KE = ½ mv2

KE: Kinetic Energy (J)m: mass (kg)v: velocity (m/s)

Example 1Which has more kinetic energy: a 1200 kg car moving at 20 km/hr or a transport truck (m=60 t) moving at 3 km/hr?

Example

1. How much energy does a 4500 kg car have if it is traveling at 35 km/hr?

2. If that same car had 1 MJ of energy, how fast is it moving?

1. How much energy does a 4500 kg car have if it is traveling at 35 km/hr?KE = ½ mv2

= 0.5 * 4500 * (35/3.6)2

= 212673 J = 2.1 x 105 J 2. If that same car had 1 MJ of energy, how fast is it

moving?KE = ½ mv2

1,000,000 = ½ * 4500 * v2

v2 = 444 m2/s2

v = 21.08 m/s = 76 km/hr

To Do…

Read pgs. 217-224Do Ch. 11 P.P. (1-8)

Skateboard Push

How do you get a person to go from zero velocity to some non-zero velocity?

You do some work!

Energy & WorkEnergy & WorkConsider an object that is accelerated from initial velocity, vo, to velocity, v, by a force,

F (net force), over a distance, d.What is the work done?

F = ma

F·d = ma·d

F·d = m(a·d) 2ad = v2 – vo2

F·d = m(v2 - vo2)

2

W = ½mv2 - ½mvo2

Kinetic Energy (energy of motion)

KE = KE = ½mv½mv22

W = KE - KEo

W = ΔKE (Work – Energy Theorem)

Conservative (Internal) vs. Non-Conservative

(External) ForcesConservative Forces Non-Conservative ForcesTotal Energy will be conserved Total Energy will not be

conserved (amount of total energy will change)

FGravity Fapplied

FSpring Force Fair

Ffriction

Work – Energy Work – Energy TheoremTheorem

The work done on an object by the net force equals The work done on an object by the net force equals the change in kinetic energy of that object.the change in kinetic energy of that object.

W = W = ΔΔKEKE

ExampleExample::A 1000 kg car going 20 m/s brakes and comes to a stop. How much work is A 1000 kg car going 20 m/s brakes and comes to a stop. How much work is done? What happens to the energy?done? What happens to the energy?

KE = KE = ½mv½mvoo22 = = ½(1000)(20)½(1000)(20)22 = = 200000 J200000 J

= = 0 – 200000 J0 – 200000 J = = ––200000 J200000 JW = W = ΔΔKEKE (lost energy)(lost energy)

Energy is dissipated as heat.Energy is dissipated as heat.

Conservation of Mechanical Energy

Total Mechanical Energy (TE):Total Mechanical Energy (TE): the sum of an object the sum of an object’’s Potential s Potential Energy (PE) and Kinetic Energy (KE). Energy (PE) and Kinetic Energy (KE).

TE = TE = PE + KEPE + KE

TE = TE = mgh + ½mvmgh + ½mv22

Law of Conservation of Energy:Law of Conservation of Energy:Within a closed (no external forces), isolated system, the Within a closed (no external forces), isolated system, the total energy of an object is conserved. total energy of an object is conserved.

EE11 = E = E22

EEinitialinitial = E = Efinalfinal

Example:

1. A pink soccer ball of mass 0.50 kg is kicked with an initial velocity of 15 m/s. Find its…

a) velocity when it is 4.0 m high.

b) height when it is going 5.0 m/s.

a)a) Einital = Efinal

mgh1 + ½mv12 = mgh2 + ½mv2

2

0 + ½(0.5)(15)2 = (0.5)(9.8)(4.0) + ½(0.5)v2

15m/sh

v

56.3 = 19.6 + 0.25v2

v = 12 m/s

b)b) 56.3 = (0.5)(9.8)h + ½(0.5)(5.0)2

h = 10 m

3.0 m3.0 m

2. Tarzan, a 75 kg Ape-man, swings from a branch 3.0 m above the ground with an initial speed of 5.0 m/s.

a) Find his velocity when he swings past ground level.

b) Find the maximum height he swings to.

c) In a separate jump, what should his initial speed be if he is to just reach a branch 4.0 m high?

Note:Note: in the physics jungle there is no air in the physics jungle there is no air resistance and the vines have no mass resistance and the vines have no mass and are unstretchable!and are unstretchable!

a)a)

b) & c)b) & c)

h = 0h = 0

a)a) mgho + ½mvo2 = mgh + ½mv2

v = √(vo2 + 2gho)

v = 9.2 m/s horizontal

b)b)

Note: The vine does NO WORK because it is always applying its force perpendicular to the direction of motion

gho + ½vo2 = 0 + ½v2

m cancels out; h = 0

v = √((5.0)2 + 2(9.8)(3.0))

h = 4.3 m

gho + ½vo2 = 0 + gh v = 0

c)c)

vo = √(2gh - 2gho)

vo = 4.4 m/s

gho + ½vo2 = 0 + gh v = 0

h = vo2 + ho

2g

3.0 m

2. Tarzan’s less famous brother, George, swings from the same 3.0 m-high branch with an initial speed of 3.5 m/s. Unfortunately, he hits a tree 1.5 m above the ground. Find his speed just before impact and the work done by the tree. His mass is 90.0 kg.

1.5 m

mgho + ½mvo2 = mgh + ½mv2

v = √(vo2 + 2gho – 2gh)

v = 6.5 m/s

gho + ½vo2 = gh + ½v2

m cancels out

v = √((3.5)2 + 2(9.8)(3.0 – 1.5))

W = ΔKE = KE - KEo

W = ½mv2 – ½mvo2

W = 0 – ½(90.0)(6.5)2

W = -1900 J

Work – Energy Theorem

The work done on an object by a net force equals the change in total energy of that object.

W = ΔTotal Energy

W = ΔKE + ΔPE

W = (KEf – KEo) + (PEf – PEo)

Question:How much heat energy is produced when a “speed” skier loses 300.0 m in elevation if he starts from rest and finishes up going 40.0 m/s? His mass (including gear) is 90.0 kg.

300m

0m/s

40m/sh = 0

Heat = Work done by frictionWork = ΔPE + ΔKE W = (PEf – PEo) + (KEf – KEo) W = (mgh – mgho) + (½mv2 – 0)

= (0 – (90)(9.8)(300)) + (½(90)(40)2 – 0)

W = -1.93 x 105 J

193000 J of heat are produced

EXAMPLE: A 3.0 kg block slides down a frictionless 58o incline with a slant length of 4.0 m. a) Assuming no friction, what is the velocity at the bottom of the plane?b) Suppose the co-efficient of friction is 0.25. What is the velocity of the block at the bottom?

EXAMPLE: A 3.0 kg block slides down a frictionless 58o incline with a slant length of 4.0 m. a) Assuming no friction, what is the velocity at the bottom of the plane?b) Suppose the co-efficient of friction is 0.25. What is the velocity of the block at the bottom?

h = 4.0*sin58o b) Work done by friction=FFd

h = 3.39 m FF = μFN = μ mgcos58o

PEtop = KEbottom =0.25(3.0)(9.8)cos58o

mgh = ½ mv2 =4.07 N(3.0)(9.8)(3.39) = 0.5(3.0)v2 W = (4.07)(-4.0) = -16.3 J (opposite

direction of motionv = 8.1 m/s W = (KEf – KEo) + (PEf – PEo)

-16.3 = 0.5(3.0)v2 – (3.0)(9.8)(3.39)v = 7.4 m/s

Homework