CONSERVATION OF ENERGY
Law of Conversation of Energy - total energy of an isolated system stays constant because it conserved over time. It cannot be destroyed or created.
Law of Conversation of Energy - total energy of an isolated system stays constant because it conserved over time. It cannot be destroyed or created.
MATHEMATICAL REPRESENTATION
Abbreviated version
Ei=Ef
Extended Version
EKi+Ugi+Usi=EKf+Ugf+Usf
THE DIFFERENT TYPES OF UNITS
SI Units of Energy
- joule
American units of energy
- calories
- BTU
NEWTON'S CRADLE EXPLAINED
Newton’s cradle shows how the ball’s energy travel through each other. The first one begins with potential energy. You hold it up in your hand and as soon as you release the ball it gains kinetic energy while it loses potential. Then, as soon as it interacts with the first ball, it squishes it, very little that you cannot tell, that is called elastic energy. The energy travels through the balls and once it gets to the other end the ball rises and gains potential energy while it loses kinetic energy. The total amount of energy is conserved the entire time.
Example problem #2:
Khaleesi is at rest with her tricycle at the top of a hill that is 8 m. How fast is she be moving at the bottom of the hill if she and the tricycle weigh 25 kg?
(mvi^2 / 2) + mghi = (mvf^2 / 2) + mgh2
(25x0) / 2 + (25 x 9.8 x 8) = (25 x v^2) / 2
1,960 = ((25 x v^2) / 2) 2
2 x 1,960 = (25 x v^2)
2 x 1,960 = (25 x v^2) / 25
(2 x 25) / 25 = v^2
((2 x 25) / 25) = √(v^2)
√((2 x 25) / 25) = v
1.41 m/s = vf
X-GAMES LOOP:
The potential energy is at its max because the skater is not moving but he is at the top of the hill.
0 m/s = vi
There is more kinetic energy here because he just finished going down the ramp and he is not going uphill.
mvi / 2 + mgh1 = mgh2 + mvf^2 / 2
mgh1 = mvf^2 / 2
gh1 = vf^2 / 2
156.8 = vf^2
√156.8 = vf
12.52 m/s= vf
There is more potential energy than kinetic energy because he is upside down at the top and there is nothing but gravity pulling him back down.
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) = ((v^2) / 2) 2
√(58.8) = √(v^2)
7.67 m/s = v
- calories
- BTU
NEWTON'S CRADLE EXPLAINED
Newton’s cradle shows how the ball’s energy travel through each other. The first one begins with potential energy. You hold it up in your hand and as soon as you release the ball it gains kinetic energy while it loses potential. Then, as soon as it interacts with the first ball, it squishes it, very little that you cannot tell, that is called elastic energy. The energy travels through the balls and once it gets to the other end the ball rises and gains potential energy while it loses kinetic energy. The total amount of energy is conserved the entire time.
Example problem #1:
An 850 kg roller-coaster is released from rest at Point A of the track shown in the figure. How fast is the roller-coaster moving at Point B? Assume there is no friction.
known:
m= 850 kg
ha= 140 m
hb= 95 m
Va= 0 m/s
Unknown:
Vb= ?
Khaleesi is at rest with her tricycle at the top of a hill that is 8 m. How fast is she be moving at the bottom of the hill if she and the tricycle weigh 25 kg?
known: unknown:
m= 25 kg vf=?
h= 8 m
g= 9.8
vi= 0 m/s^2
g= 9.8
vi= 0 m/s^2
(mvi^2 / 2) + mghi = (mvf^2 / 2) + mgh2
(25x0) / 2 + (25 x 9.8 x 8) = (25 x v^2) / 2
1,960 = ((25 x v^2) / 2) 2
2 x 1,960 = (25 x v^2)
2 x 1,960 = (25 x v^2) / 25
(2 x 25) / 25 = v^2
((2 x 25) / 25) = √(v^2)
√((2 x 25) / 25) = v
1.41 m/s = vf
X-GAMES LOOP:
0 m/s = vi
mvi / 2 + mgh1 = mgh2 + mvf^2 / 2
mgh1 = mvf^2 / 2
gh1 = vf^2 / 2
(9.8) (8) = vf^2 / 2
(78.4) 2 = (vf^2 / 2) 2156.8 = vf^2
√156.8 = vf
12.52 m/s= vf
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) = ((v^2) / 2) 2
√(58.8) = √(v^2)
7.67 m/s = v