Transcript Conservation of Energy Notes File

Conservation of Energy
Total Mechanical Energy
• The total amount of mechanical energy is the
sum of the potential energy and the kinetic
energy.
• Mechanical energy can be either kinetic
energy (energy of motion) or potential energy
(stored energy of position).
• For example, a moving car possesses
mechanical energy due to its motion (kinetic
energy).
• A moving baseball possesses mechanical
energy due to both its high speed (kinetic
energy) and its vertical position above the
ground (gravitational potential energy).
Mechanical Energy
• The total amount of mechanical energy in a
CLOSED system will remain constant
Physical System
• A defined collection of objects
Closed System
• A collection of objects that does not lose or
gain mass
Law of Conservation of Energy
Law of Conservation of Energy
• When a ball flies through
the air it does not lose
energy even though it
slows down and speeds
up
• The energy in the ball
changes from kinetic to
potential and then back
to kinetic energy
• When an object moves down the track, kinetic
energy______ and potential energy _______.
• When an object moves up the track, kinetic
energy______ and potential energy _______.
Where is PE the greatest?
Where is PE the lowest?
Where is KE the highest?
Where is KE the lowest?
Mechanical Energy
• The total amount of mechanical energy is the
sum of the potential energy and the kinetic
energy.
ME = KE + PE
ME = ½mv² + mgh
Acceleration due to gravity
g = 9.8 m/s/s ~ 10m/s/s
g= 10m/s/s
• A 0.3kg softball is thrown straight up in the air
with a kinetic energy of 50J. If the ball is 2m
above the ground when it is released from the
thrower’s hand, what is the total mechanical
energy of the ball at the point of release?
TME = KE + PE
TME = ½mv² + mgh
• A 0.3kg softball is thrown straight up in the air
with a kinetic energy of 50J. If the ball is 2m
above the ground when it is released from the
thrower’s hand, what is the total mechanical
energy of the ball at the point of release?
TME = KE + PE
TME = ½mv² + mgh
TME= 50 + (0.3 * 10 * 2)
TME= 56J
• A 10.00-kg model plane flies horizontally at a
constant speed of 15.50 m/s. The plane is
25.00 meters above the ground. The total
mechanical energy of the plane is ______.
TME = KE + PE
TME = ½mv² + mgh
• A 10.00-kg model plane flies horizontally at a
constant speed of 15.50 m/s. The plane is
25.00 meters above the ground. The total
mechanical energy of the plane is ______.
TME = KE + PE
TME = ½mv² + mgh
TME= (½ * 10 * 15.5²) + (10*10*25)
TME = 1201.25 + 2500
3701.25J
Equations
If
MEi = MEf
Then
KEi + PEi = KEf + PEf
And
½mvi² + mghi = ½mvf² + mghf
The first hill of the Titan roller coaster at Six Flags has a height
of 80 m. The car at the top has a mass of 1200 kg and is
initially at rest. Find the kinetic energy of the roller coaster car
at the bottom of the hill.
h= 80m
m= 1200kg
vi= 0  KEi = 0
hf= 0  PEf = 0
KEi + PEi = KEf + PEf
0 + PEi = KEf + 0
PEi = KEf
PEi = KEf
mghi = KEf
1200*9.8*80 = KEf
940800 J = KEf
The first hill of the Titan roller coaster at Six Flags has a
height of 80 m. The car at the top has a mass of 1200 kg and
is initially at rest.
Find the velocity of the roller coaster car at the bottom of the
hill.
h= 80m
m= 1200kg
vi= 0  KEi = 0
hf= 0  PEf = 0
KEi + PEi = KEf + PEf
0 + PEi = KEf + 0
PEi = Kef
Therefore:
mghi = ½mvf²
mghi = ½mvf²
1200*9.8*80 = ½*1200*vf²
940800 = 600*vf²
1568 = vf²
39.6 m/s = vf