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Reading Segment:
Conservation ofMechanical Energy
To prepare for this section, please read:
Unit 4: p. 14
D. Conservation of Energy
Recall:
There are two major advantages to the energetics perspective:
1. Energy is a scalar
- does not have direction
- no need for vector math
2. It is conserved
We will now analyze situations where total mechanical energy is conserved.
Conservation of Mechanical Energy
- total mechanical energy stays constant (conserved) when:
* the system is closed
- no objects are added / lost
* no friction
* only mechanical forms of energy change
- only Fg and Fs do work on the object
- mechanical energy can convert only to another
form of mechanical energy
Equations:
If mechanical energy is conserved, then you can use two equations:
1. EmT remains constant
EmTi = EmTf
Epgi + Eki = Epgf + Ekf
2. No overall change in energy
Epg + Ek = 0
Epg = - Ek
That is,
if Epg increases by 100 J,
then Ek decreases by 100 J (no overall change)
So, one form of mechanical energy transforms into
another form of mechanical energy
e.g. Consider an object thrown downward (No air resistance)
Epgi = 600 J
Eki = 200 J
Epgf Ref h (h = 0)
Ekf
What is the total mechanical energy at the start?
EmTi Epgi = 600 J
800 J Eki = 200 J
Epgf Ref h (h = 0)
Ekf
What is the total mechanical energy at the end?
EmTi Epgi = 600 J
800 J Eki = 200 J
EmTi = EmTf
EmTf Epgf Ref h (h = 0)
800 J Ekf
So, what is its kinetic energy at the end?
Epgi = 600 J
Eki = 200 J
Calculate and interpret:
- the change in Epg
- the change in Ek
Epgf = 0
Ekf = 800 J
Epgi = 600 J
Eki = 200 J Epg = Epgf - Epgi
= 0 - 600 J
= - 600 J
object loses 600 J of Epg
Epgf = 0 (since it loses height)
Ekf = 800 J
Epgi = 600 J
Eki = 200 J Ek = Ekf - Eki
= 800 J - 200 J
= + 600 J
object gains 600 J of Ek
Epgf = 0 (since it gains speed)
Ekf = 800 J
Epgi = 600 J
Eki = 200 J
The object loses 600 J of Epg,
but it gains 600 J of Ek
We say that the Epg has
transformed into Ek.
Epgf = 0
Ekf = 800 J
Animations:
1. Dropped object:
http://departments.weber.edu/physics/amiri/director/dcrfiles/energy/FallingBallS.dcr
Can you explain what happens in the animation?
You should have noticed:
The Epg goes down by 1078 J,
but at the same time, the Ek goes up by 1078 J
That is, the Epg has transformed entirely into Ek
Thus, the total mechanical energy will remain the same
throughout the entire motion
Other animations showing conservation of energy:
2. Pendulum:
http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t
=27
3. Rollercoaster:
http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/qt/energy/coastwin.html
Ex. A ball is dropped from a height H and it lands with a speed of 7.0 m/s. If the system is conservative,
a) find H
b) sketch a Epg vs t, Ek vs t, and a EmT vs t graph
a) EmTi = EmTf Rest
Epgi + Eki = Epgf + Ekf
Ref h (h = 0)
Be certain to state 7.0 m/s
where the height is zero
a) EmTi = EmTf Rest
Epgi + Eki = Epgf + Ekf
Ref h (h = 0)
7.0 m/s
Since it starts at rest, Eki = 0
Since it has no height at the end, Epgf = 0
EmTi = EmTf
Epgi + Eki = Epgf + Ekf
mghi = 0.5 mvf2
ghi = 0.5 vf2
If you divide both sides of the equation by m,
the mass cancels
Thus, the answer does not depend on mass.
i.e. It is true for any mass
EmTi = EmTf
Epgi + Eki = Epgf + Ekf
mghi = 0.5 mvf2
ghi = 0.5 vf2
hi = 0.5 vf2 = 0.5 (7.0 m/s)2
g 9.81 m/s2
H = 2.5 m
Ex. 3 A 1.20 kg car travels the following path. No friction.
A 6.70 m/s
v ?
C
8.70 m B
4.20 m
1.90 m
D
Find its speed at C.
A 6.70 m/s
v ?
C
8.70 m B
4.20 m
1.90 m
D (h = 0)
The reference height (h = 0) is at D,
the lowest location in the diagram
A 6.70 m/s
v ?
C
8.70 m B
4.20 m
1.90 m
D (h = 0)
Total mechanical energy remains the same.
So, EmTA = EmTC
EpgA + EkA = EpgC + EkC
Ref h = D
EpgA = mghA
= (1.20 kg) (9.81 N/kg) (8.70 m) = 102.42 J
EkA = 0.5mvA2
= 0.5 (1.20 kg) (6.70 m/s)2 = 26.93 J
EpgC = mghC
= (1.20 kg) (9.81 N/kg) (4.20 m) = 49.44 J
Ex. 4
A 14.0 kg object is currently moving at a height of 75.0 cm. If it then loses 92.0 J of Ek, find the object's new height.
Assume a conservative system.
Solution
The total mechanical energy must remain the same
So, if the object loses 92.0 J of Ek,
it must gain 92.0 J of Epg
That is, the Ek converts (transforms) into Epg