2. Definition
The total amount of energy of an isolated system is conserved/ constant
3. Mathematical Representation
Abbreviated version: Ef= Ei
Extended Version: KEi+PEi+OEi=KEf+PEf+OEf
4. Units of Energy
SI Units: Joules
American Units: Calories, horsepower
5. Video
The video talks about how a roller coaster conserves energy. As the video goes on it explains how the chains help the roller reach the top of the hill to gain potential energy. No engine is helping the roller coaster go to the top of the hill. As the roller coaster falls the potential energy transforms into kinetic energy. It also mentioned that when the roller coaster is ready to fall the only force that relies on is the force of gravity but only when it falls. This makes the roller coaster speed up when it is near the ground. As soon as the roller coaster goes to another loop it slows down and now the kinetic energy transforms to potential energy. It will continue to do that until the roller coaster stops.
6. Problem #1
During a flood a tree trunk of mass of 100 kg falls down a waterfall. The waterfall is 5 m high.
If air resistance is ignored, calculate:
- The potential energy of the tree trunk at the top of the waterfall.
- The kinetic energy of the tree trunk at the bottom of the waterfall.
- The magnitude of the velocity of the tree trunk at the bottom of the waterfall.
m= 100 kg
height= 5 m
Potential energy at the top
Kinetic Energy at the bottom
Velocity at the bottom
Calculation for Potential Energy
PE= mgh
=(100 kg) (9.8 m/s^-2) (5 m)
=4,900 J
Calculation for kinetic Energy
KE
PE = KE
PE = KE
The kinetic energy of the tree trunk at the bottom of the waterfall is equal to the potential energy it had at the top of the waterfall
Therefore KE= 4 900J
Calculation for Velocity
KE= 1/2 m(v)^2
4 900= 1/2(100 kg) (v^2)
98= v^2
v=-9.899
v= -9.90
Problem #2
A boy with a mass of 40 kg is on the top of a tree that is about 15 m high he is trying to jump on a bunch of pillows for a YouTube video. What is his final velocity?
Known:.
M= 40 kg
h= 15 m
g= 9.8 m/s^2
Vi= 0 m/s
Unknown:
Vf
Solution
mv^2 / 2i + (mgh)i=mv^2 / 2f + mghf
0+ 5880 = 40 v^2 + 40 (-9.8) (12)
2(5880)= 40(vf ^2) / 2
x2
11760/ 40 = 40 (v^2) / 40
√294 = √v^2
v= 17.146 m/s
mv^2 / 2i + (mgh)i=mv^2 / 2f + mghf
0+ 5880 = 40 v^2 + 40 (-9.8) (12)
2(5880)= 40(vf ^2) / 2
x2
11760/ 40 = 40 (v^2) / 40
√294 = √v^2
v= 17.146 m/s
X Games Loop
- When he is at the fastest his potential energy transforms to kinetic energy.
2. When he starts at the top he only has potential
energy and 0 kinetic energy. The graph looks like
that because he is up high.
3. He is upside down so the kinetic energy he had
transforms to potential energy because the skater
slows down.
Calculations
1. Fastest
mgh = mvf ^2 /2
9.8 ( 8 ) = vf^2 /2
2(78.4) = vf^2
√156.8 = √vf^2
v=12.52 m/ s
2. Max height
v= 0 m/ s
3. Top of Loop
mgh1 = mgh2 + mv^2/2
gh1- gh2 = v^2/2
9.8(8) - 9.8 (5) = v^2 /2
78.4 - 49 = v^2/2
2(29.4) = 2(v^2)
√58.8 = √v^2
v = 7.668 m/s