The Law of Conservation of Energy

2. Definition: the law of Conservation of Energy states that the total energy of an isolated system remains constant

3. Mathematical Representation for the Law of Conservation
Abbreviated version: Ei=Ef 
Extended Version: GPEi + SPEi + KEi = PGF + SPEi + KEi

4. Different Units of Energy
SI units: Joules: -Joules (Watt SI units for energy)
American Units: -British Thermal Unit (BTS) -Calories (cal)

5. Link to a Video
Visual Of Conservation Of Energy
In this video it shows how roller coasters conserves energy. At the start the roller coaster has stored potential energy at the beginning. It also has a constant kinetic energy since its moving. A few seconds later the roller coaster starts to go up and its potential energy increases because its moving up. While the cart is going up its kinetic energy decreases same reason because its going up. After a couple seconds the cart goes on a flat rail for a short time, then the cart goes down in height and losses potential energy. its kinetic energy increases since its going faster at a downwards motion.

Sample problem 1
6. The diagram below shows a 10,000 kg bus traveling on a straight road which rises and falls. The horizontal dimension has been foreshortened. The speed of the bus at point A is 26.82 m/s (60 mph). The engine has been disengaged and the bus is coasting. Friction and air resistance are assumed negligible. The numbers on the left show the altitude above sea level in meters. The letters A–F correspond to points on the road at these altitudes.

1. Find the speed of the bus at point B.
2. An extortionist has planted a bomb on the bus. If the speed of the bus falls below 22.35 m/s (50 mph) the bomb will explode. Will the speed of the bus fall below this value and explode? If you feel the bus will explode, identify the interval in which this occurs.
3. Derive an equation to determine the speed of the bus at any altitude.

Sample Problem 2

Elmer has a mass of 22 kg he is standing on a 5 meter cabin. He decides to jump off 

into the pond. How fast is he moving when he hits the water.

Known

M=22kg

H=5m

G=9.8m/s^2

Vi=0

Unknown

Vf=?

(mv^2/2) initial+mgh initial (mv^2/2)final

+mgh final

(22x0)^2/2+(22)(9.8)(5)=(22xv^2/2)+22

                                        (9.8)(5m)

(22x0)^2/2+(22)(9.8)(5)=(22xv^2/2)

1078=(22xv^2/2)

           1078=(22xv^2/2)

x2            x2

2(1078)=22xv^2

2(1078)=22xv^2

22              22

√98=√v^2
v=9.89m/s

7. 

This is the first image. The skater is at his highest potential energy because he is at the highest point of the ramp. Of course the kinetic energy is low because he has not begun to speed up.

calculations = 0m/s

This is the second image potential is still the highest because he is still pretty high up in the ramp. The kinetic however is still low because he slows down. But as soon as he makes it around the loop the kinetic will be the highest because he will gain speed.

calculations

mgh1=mgh2+mv^2/2

gh1=gh2=v^2/2

gh1-gh2=v^2/2

√2(gh1-gh2)=√v^2

√2(gh1-gh2)=v

√2(9.8)(8)-(9.8)(6)=v

6.26m/s=v

This is the third image the skater has made his way down the ramp and kinetic energy is higher than potential. Since the skater is going down he gains some serious speed thats why kinetic surpasses potential energy.

mvi/2+mgh1=mvf^2/2=mgh2

mgh1=mvf^2/2

gh1=vf^2/2

9.8(8)=vf^2/2

2(78.4)=vf^2

156.8=vf

√156.8=√vf^2

12.52m/s=v