Mass and Energy

W = workif the work done to increase kinetic energy then

W = ΔEk = Ek2 - Ek1

= 0.5mv22 - 0.5mv1

2

= 0.5m(v22 - v1

2)

•Note: For classical mechanics we assume that mass is constant and that all energy would be transferred as speed.

•(classical mechanics can be used if v is less than 0.1c)

•This is not true as speeds become relativistic, close to the speed of light, v at or greater than 0.1c.

•Einstein suggested that the total relativistic energy with a rest mass of m, moving at a speed of v, relative to an inertial frame is:

Etotal = mc2/(1- v2/c2) 0.5

where Etotal = total mass energy (in J)m = rest mass (in kg)v = speed of an object (in m/s)

•If at rest in at inertial frame then v = 0

Therefore Etotal=mc2/(1- (0)2/c2)0.5

= mc2/(1- 0)0.5 = mc2/(1)0.5

= mc2/1 = mc2

Thus Erest = mc2

Where Erest = rest mass energy (in J)

Einstein proposed:1. rest mass is a form of energy which all objects with mass have.

2. there might be forces or interactions in nature that transform mass energy into other types of energy

Energy Mass Equivalence

•Also implies that as you acquire energy, you are also acquiring mass▫The faster you move, the heavier you are▫This is why we need relativistic momentum

•According to this equivalence,▫Light has mass because it has energy▫But light has zero rest mass

•In classical mechanics; mass and energy are conserved separately.

•Conservation of Mass-Energy: the principle that rest mass and energy are equivalent.

Etotal = Erest + Ek

Ek = Etotal - Erest

= mc2/(1- v2/c2)0.5 - mc2

= mc2[1/(1- v2/c2)0.5 -1]

Sample Problem # 1

A 4.0 kg of coal is burned. The thermal energy released from the combustion of coal is about 3.3 X 107 J/kg. What is the % efficiency of the burning of the coal?

•Solution Etotal = mc2

Ein = mc2

= (4.0 kg)(3.00 X 108 m/s)2

= 3.6 X 1017 J

Eout = (4.0 kg)(3.3 X 107 J/kg)= 1.32 X 108 J

% eff = Eout X 100% Ein

= 1.32 X 108 J X 100% 3.6 X 1017 J = 3.7 X 10-8 %

The % efficiency of coal is 3.7 X 10-8 %.(Coal is not very efficient!!)

Example # 2: Find the rest mass energy of a 1.0 kg object

Solution Erest= mc2

= (1.0 kg)(3.00 X 108m/s)2

= 9.0 X 1016 JThe rest mass energy of a 1.0 kg object is 9.0 X 1016 J.

Example # 3: An proton is moving at about 0.9c. Knowing 1 eV = 1.602 X 10-19 J, calculate the protons's:

(a) rest energy in MeV.(b) total energy in MeV.(c ) kinetic energy in MeV.Mproton = 1.67x10-27kg

Solution (a) Erest = mc2

=(1.67X10-27kg)(3.00X108m/s)2

=8.199X10-14JX(1eV/1.602X10-19J)= 5.12 X 105 eV= 0.512 MeVThe rest energy is 0.512 MeV.

(b) Etotal = mc2/(1- v2/c2) 0.5

= (9.11X10-31kg)(3.00X108m/s)2

[1-(0.90c)2/c2] 0.5

=1.88097X10-13J X(1eV/1.602 X10-

19J)= 1.17 X 106 eV= 1.17 MeVThe total energy is 1.17 MeV.

(c ) Etotal = Erest + Ek

Ek = Etotal - Erest

= 1.17 MeV – 0.512 MeV= 1.11 MeVThe kinetic energy of the

electron is 1.11 MeV.

•"In a nuclear process, you convert something like one part in 1,000 of your rest mass into energy, whereas if you fall into a black hole, you can convert something like 20, 30, 40 percent," Hogg says. "So from the point of view of the energetics of the universe, these black holes are important, because they are big converters of rest mass into energy."

11.3 Practice Questions

•Page 583 Questions 1-6 •Page 584 Question 3

Chapter 11 Review Questions

•Page 588 Questions 1-14, 16-19, 22•Page 590 Questions 4-13, 27

Einstein cartoons

Ok,so this is not a cartoon.

Also not a cartoon, Einstein spent the last of his days trying to find a theory that unifies all types of fields. The String Theory is the most commonly accepted unifying theory, but it is far from complete.

ALSO NOT EINSTEIN CARTOONS Incompetent Incompetent

Slug Slug AssassinsAssassins

May I be May I be excused my excused my brain is full.brain is full.

ALSO NOT EINSTEIN CARTOONS

ALSO NOT EINSTEIN CARTOONS

ALSO NOT EINSTEIN CARTOONS

ALSO NOT EINSTEIN CARTOONS