Download CHAPTER 8- POTENTIAL ENERGY and CONSERVATION of

CHAPTER 8- POTENTIAL ENERGY and CONSERVATION of ENERGY
POTENTIAL ENERGY
Potential Emergy is energy stored by thepositioning of an object in a force field or by the
positioning of a mechanical object.
y
U = m g (y-yo)
Fg
gravity
dy
yo
U = 1/2 k ∆x2
spring
Fs
∆x
dx
In each case an increase in potential energy ∆U is associated with an amount of work
W=-
∫
F dx where the negative sign implies the force and displacement are in opposite
directions.
CONSERVATION OF ENERGY
Since a positive change in kinetic energy ∆K is associated with positive work we can say
potential and kinetic energy work oppositely, that is
W = ∆K = - ∆U
or
∆K + ∆U = 0
If we define an objects Total Energy as E = K+ U then ∆E = 0 along the path of any
object acted upon by a force. Any loss of energy due to frictional forces Wf must be
included.
EB
EA
Wf
EA = EB + | Wf |
∆EBA = EB - EA = | Wf |
SUPER MARIO (No Friction)
If a spring-mass (Super Mario in class ) is compressed by ∆y and released how high will
it jump above the table?
E(bot) = 1/2 k ∆y2 (starting from rest K = 0)
E(top) = m g h
(At top K = 0)
Since energy is conserved
E(bot) = E(top)
1/2 k ∆y2 = m g h
h = ( 1/2 k ∆y2 )/ mg
∆y
h
INCLINED PLANE
A mass M slides down an inclined plane from a height H. What is its velocity at the
bottom?
From conservation of energy over the
frictionless surface we can write
H
E(bot) = E(top)
1/2 m V2 = m g H
V= (2gH)
θ
V
1/2
INCLINED PLANE with friction
A mass M slides down an inclined plane (coeficient of kinetic friction µK) from a height
H. What is its velocity at the bottom?
From conservation of energy over the
frictionless surface we can write
H
E(bot) = E(top) - |Wf | = E(top) - µ M g sin(θ) L
1/2 M V2 = M g H - µ M g sin(θ) (H / sin(θ) )
V = ( 2 g H - 2 µ g H tan(θ) ) 1/2
L
θ
V
LOOP-THE-LOOP
A mass M ris eleased from a height 4R and enters a circular loop-the-loop of radius R.
EA = M g 4R
EB = 1/2 M VB2 EC = 1/2 M VC2 + M g R
A
ED = 1/2 M VD2 + M g 2R
MV2/R
N
g
Conservation of energy
EA = EB = EC = ED
Mg
D
4R
C
B
Does the mass M stay on the track at D or fall down?
Force balance at D gives:
N + Mg =MVD2 / R
If N=0 then the mass may lift-off.
Mg < MVD2 / R is the condition for falling such that VD is not large enough.
GLOBAL ENERGY CONSERVATION
Even though we defined cases where energy was not conserved due to frictional forces
for example, if we could account for theenergy loss we could make the statement that
“global energy” or “total energy” was always conserved! Scientist believe that the “total
energy” of the universe will remain the same even though it may change its form from
electrical, mechanical, gravity, frictional, heat, atomic, nuclear, etc.
E = K + Ue + Umech + Ug + Ufrict + Uheat + Uatomic + Unucl = Universal Constant
EINSTEIN’S ENERGY
Einstein defined energy in relativistic terms. He counted the mass Mo of an object as
part of its energy. This shocked the world of classical physics!
E = M c2 = K + U + Mo c2
He realized that if nucleus A split in to B and C and MA > MB + MC
or ∆M= MA –( MB + MC) and excess of energy is released E = ∆M c2
This excess of energy comes from the nuclear binding potential Unuc
If a uranium atom releases E=10-10 J and there are 6.02 x 1021 atoms
in a mole then Etot = 6 x 1011 J of energy released !