Passing This Class

2)The law of  Conservation is that it's of an isolated system remain constant 


 3) Ei=Ef is the Short Version

Extended Version is KEi+PEi+WnC+OEi=KEf+PEf+OEf

4) SI Units: -Joules
 -watt (SI units for Electrons Energy In U.S)
American Units: -Foot-pound(ft x lb)
-British Thermal Units (BTU)
-Calories

5) Unique video: https://www.youtube.com/watch?v=LrRdKmjhOgw

In the video the roller coaster  shows that when the roller coaster has a large quantity of potenial energy when it's at the top of the hill. The kinetic energy is energy of the motion, specifically velocity. When the car is at the bottom of the roller coaster going through the dip, it has it's max kinetic energy, in between top to bottom of the roller coaster. When the car is going up or coming down that is where potential energy and kinetic energy trades off.

6a) A cart travels along a frictionless roller coaster track. At point A, the cart is 10 m above the ground and traveling at 2 m/s.

A) What is the velocity at point B when the cart reaches the ground?
B) What is the velocity of the cart at point C when the cart reaches a height of 3 m?
C) What is the maximum height the cart can reach before the cart stops?

6 b)During the hurricane a large tree with a mass of 22.0 kg and height of 13.3 m above ground falls on a roof that is 6.0 m above ground

a. forget the air resistance, find the kinetic energy of the limp when it reaches the roof.

b. what is the speed of the limp when it reaches the roof

1) known: m=22 g=9.8 m/s^2  H limp=13.3  Vi=0.0  Hroof=6m KEi=0.0j

2)unknown: PEi=? KEi=? PEf=? Vf=?



solve for the internal height of the limb relative to the roof
H=Hlimp - Hroof
H=13.3m-6m
H=7.3m

Sole the initial potential energy of the limb
PEi=mgh
PEi= (22kg)(9.8m/s^2)(7.3m)
KEf=1.6x10^3j or 1,600j

Identify the kinetic energy of the limb
KEf=0j

The kinetic energy of the limb when it reaches the roof is equal to the initial potential of energy because is stored.
KEf=PEi
=1.6x10^3j or 1,600j

Solving the speed of the limb
KEf=1/2(m)vf^2
Vf=square root of 2KEf/m
Vf=square root of 2(1,600j)/22k
Vf=12m/s

 7 A)

B) firstly, the one thing that stays the same is the total energy.In the first picture when the skater is at the top of the loop he has the highest potential energy.
In the second picture when the skater is at the top of the loop he has the lowest kinetic energy and medium potential energy. In the third picture when he’s at the start of the loop he has the highest kinetic energy.

8) calculation

fastest
mgh=mvf^2/2
9.8x8=vf^2/2
2(78.4)=vf^2
square root (156.8)
v=12.52m/s

The Max Height
v=0m/s

The Top loop
mgh1=mgh2+mv^2/2
9.8(8)-(9.8)5=v^2/2
2(29.8)=59.6
square root 59.6=
7.72m/s