Conservation of Energy
Law of Conservation of Energy - energy cannot be created nor destroyed but it can be changed from one form to another.
Mathematical Representation:
Abbreviated Version
Ei = Et
Extended Version
GPEi + SPEi + KEj = PGEf + SPEf + KEf
Different Units of Energy:
SI units:
Joule (J)
American units:
calorie (cal) , BTU
Video: https://www.youtube.com/watch?v=LrRdKmjhOgw
In the video the roller coaster is gaining potential energy when it gets to the top of the hill. once the cart was released from the top of the hill it converted to kinetic energy increasing it's velocity. The total energy remains constant throughout the ride. In the roller coaster the energy is constantly being transferred between kinetic and potential, only during motion.
Example Problem #1 : Tony Hawk
rides his skateboard at a local skate park. He starts from rest at the top of the track as seen in the figure below and begins a descent down the track, always maintaining contact with the surface. The mass of the skateboard is negligible, as is friction except where noted.
Example problem #2: Chelsy is riding her skateboard at 12m/s but decides to ride down a hill of 10m. How fast is she moving at the bottom is chelsy weighs 50kg?
the potential energy is higher then the kinetic energy because the cart is not in speed and gravity trys to pull it back down.
known: uknown:
h1= 8m Vf= ?
h2= 5m
g= 9.8
mgh1 = mgh2 + mVf^2/2
gh1 = gh2 + mVf^2/2
gh1 - gh2 = gh2 - gh2 + m Vf^2/2
(9.8) (8) - (9.8) (5) = (Vf^2) / 2
78.4 - 49 = (Vf^2) / 2
2 x (29.4) = (Vf^2) / 2) x 2
√(58.8) = √(Vf ^2)
Vf = 7.67 m/s
Example Problem #1 : Tony Hawk
rides his skateboard at a local skate park. He starts from rest at the top of the track as seen in the figure below and begins a descent down the track, always maintaining contact with the surface. The mass of the skateboard is negligible, as is friction except where noted.
(a) What is Tony's speed when he reaches the bottom of the initial dip, 18.0 m below the starting point?
(b) He then ascends the other side of the dip to the top of a hill, 8.0 m above the ground. What is his speed when he reaches this point?
Solution:
a. From the conservation of energy: Potential energy at the top of the 18 m transforms into the Kinetic energy at the bottom of the dip.
b. From the conservation of energy: Potential energy at the top of the 18 m transforms into the Kinetic and Potential energy at the top of a hill.
Answer 
and 
and Example problem #2: Chelsy is riding her skateboard at 12m/s but decides to ride down a hill of 10m. How fast is she moving at the bottom is chelsy weighs 50kg?
known: unknown:
Vi= 12m/s Vf= ?
Vi= 12m/s Vf= ?
h= 10m
m= 50kg
g= 9.8
(mVi^2/2) + mgh1 = (mVf^2/2) + mgh2
(50 x 12^2)/2 + (50 x 9.8x10) = (50 x Vf^2)/2
8,500 = (50 x Vf^2)/2) x 2
2 x 8,500 = (50 x Vf^2)
(2 x 8,500)/50 = (50 x Vf^2)/50
√(2 x 8,500)/50 =√Vf^2
√(2 x 8,500)/50 = Vf
Vf = 1.02m/s
8,500 = (50 x Vf^2)/2) x 2
2 x 8,500 = (50 x Vf^2)
(2 x 8,500)/50 = (50 x Vf^2)/50
√(2 x 8,500)/50 =√Vf^2
√(2 x 8,500)/50 = Vf
Vf = 1.02m/s
X Games Loops:
The potential energy is at its greatest because it is stored up due to the skater not moving at the top of the hill. It is at 0m/s
h= 8m
h= 8m
known: uknown:
h1= 8m Vf= ?
h2= 5m
g= 9.8
mgh1 = mgh2 + mVf^2/2
gh1 = gh2 + mVf^2/2
gh1 - gh2 = gh2 - gh2 + m Vf^2/2
(9.8) (8) - (9.8) (5) = (Vf^2) / 2
78.4 - 49 = (Vf^2) / 2
2 x (29.4) = (Vf^2) / 2) x 2
√(58.8) = √(Vf ^2)
Vf = 7.67 m/s
Kinetic energy is higher here because the skater finishes the loop giving him more speed while he's moving.
(mVi^2/2) + mgh1 = (mVf^2/2) + mgh2
gh1 = Vf^2 / 2 known:
(9.8) (8) = Vf^2 / 2 h = 8
(78.4) x 2 = (Vf^2 / 2) x 2 g = 9.8
√156.8 = √Vf^2 unknown:
√156.8 = Vf Vf = ?
Vf = 12.52 m/s
Vf = 12.52 m/s