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Aim: How can we explain the Aim: How can we explain the Law of Conservation of Energy?Law of Conservation of Energy?
Do Now:
Homework Review
Conservation of EnergyConservation of Energy
The total energy (EThe total energy (ETT) of an ) of an
object or system is constantobject or system is constantEnergy cannot be created or Energy cannot be created or
destroyeddestroyedEnergy Energy cancan be changed from be changed from
one form to another.one form to another.
A Dropped SphereA Dropped SphereWhat energy does it have when held What energy does it have when held
above the ground?above the ground?
Potential EnergyPotential EnergyWhat happens to this PE as the ball What happens to this PE as the ball
drops?drops?
It becomes smaller and smallerIt becomes smaller and smallerIs the energy just disappearing?Is the energy just disappearing?
No!! It is being converted into KENo!! It is being converted into KE
(the object is speeding up)(the object is speeding up)
100 m
m = 10 kg
At the top:
PE = 9,800 J (PE = mgh)
KE = 0 J (at rest)
ET = 9,800 J (PE + KE = ET)
At the bottom:
PE = 0 J (no height)
ET = 9,800 J (total energy is constant!)
KE = 9,800 J (PE + KE = ET)
If PE = 4,900 J, what is KE?
KE = 4,900 J
PE = 3,000 J, what is KE?
KE = 6,800 J
As an object falls, PE is being converted into KE
ΔPE = ΔKE
mgΔh = ½mv2
Problem
A 2 kg mass is dropped from a height of 10 m. What is the KE as it strikes the ground?
ΔKE = ΔPE
ΔKE = mgΔh
ΔKE = (2 kg)(9.8 m/s2)(10 m)
ΔKE = 196 J
How fast is the object moving when it strikes the ground?
KE = ½mv2
196 J = .5(2kg)v2
196 J/kg = v2
v = 14 m/s
After the mass falls 5 m, what is its KE?
ΔKE = ΔPE
ΔKE = mgΔh
ΔKE = (2 kg)(9.8 m/s2)(5 m)
ΔKE = 98 J
What is its PE?
Solution 1
PE = mgh
PE = (2 kg)(9.8 m/s2)(5 m)
PE = 98 J
Solution 2
PE + KE = ET
PE + 98 J = 196 J
PE = 98 J
Sometimes energy is lost due to heat or friction
When this happens:
PE + KE + Q = ET
Q = energy lost due to friction
It is not used in every problem!
10 m
m = 2 kg
KE = 190 J
How much energy was lost due to friction?
This means solve for Q
At the top:
•No KE
•Has not moved, so no frictional loss
•The only energy is PE
•PE at the top equals ET
PE = mgh
PE = (2 kg)(9.8 m/s2)(10 m)
PE = 196 J
•At the bottom, there is no height (no PE)
•KE should equal PE at the top, unless there is frictional loss
PEtop = 196 J
KEbottom = 190 J
PE + KE + Q = ET
0 J + 190 J + Q = 196 J
Q = 6 J