# Conservation of Mechanical Energy Ef S Ei What if all forces are

advertisement ```Analyses with Mechanical Energy
ENERGY METHODS
Learning Objectives
After you complete the homework associated with this
lecture, you should be able to:
• Be able to draw an energy diagram to explain the
changes of speed and direction of motion of an object
whose mechanical energy is constant.
• Use energy methods to analyze motion in systems where
work is done by forces other than those that produce
potential energies included in mechanical energy.
• Describe how total energy is conserved even when
dissipative forces act.
The Mechanical Energy ( E ) of a system is the sum of its
Kinetic (K ) and Potential (U ) energies
Emechanical / K + U
This gives us a VERY useful tool to analyze Nature !
Ef – Ei = (Wother)i6f
The change in the total mechanical energy of a system of
particles in going from the initial to the final state equals the
work done on the system by forces other than those represented
through potential energy.
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Conservation of Mechanical Energy
Motion in a Potential Energy Well
Ef S Ei = (Wother)i6f
What if all forces are accounted for in potential energies?
Then Wother= 0, and
Ei = Ef if (Wother)i6f = 0
Mechanical energy is conserved if no non-conservative
forces acts, i.e., if only conservative forces are present
(say P
Fa , P
F b , ... producing potential energies Ua ,Ub , ....)
Ei = &frac12; m1 v1i2 + &frac12; m2 v2i2 + &middot;&middot;&middot; + Uai + Ubi + &middot;&middot;&middot;
= Ef = &frac12; m1 v1f2 + &frac12; m2 v2f2 + &middot;&middot;&middot; + Uaf + Ubf + &middot;&middot;&middot;
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Consider a one-dimensional potential well .
Fx(x) = S dU/dx
Ef = Ei = E = constant
K(x) + U(x) = E
= constant
K(x) = E S U(x) gives
variation of K with x.
Potential U(x) low
Y Kinetic K(x) is high
Potential U(x) high Y Kinetic K(x) is low
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Turning points xt are places where object stops and turns
around where K(xt) = 0 , i.e., where E = 0 + U(xt) .
BALL ON DOUBLE-INCLINED PLANE
'
C
Little friction Y
Ef S Ei = W other Y Ei = Ef
C
Ei = Ki + Ui = &frac12; m '
v i2 + mgyi = 0 + mg(0) = 0 !
How high does the ball rise? Since K = &frac12; m v 2 \$ 0,
U can only increase until Kf = 0.
Thus ball rises to state such that Ef = 0 + mgyf .
Ef = Ei = 0
Y
yf = yi = 0 Y rises to same height.
: Trust-in-the-Physics Pendulum
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When Other Forces Act ...
Example of Using ΔE = Wother
What happens if there are other forces present that are nonconservative or for which you don’t know their potential form?
Then just use the good ol’ all-purpose energy analyzing tool:
A rocket at rest is ignited on top of a frictionless cliff. It has a
constant horizontal thrust force of T. After traveling distance
L along the flat top, it leaves the cliff. When it arrives a
distance d from the cliff face at point P, it is a distance h below
the cliff top. Its velocity is angle θ down from the horizontal.
Assume its mass m is constant. What is its speed V at P ?
Question to ask yourself: What toolbox do we use?
Ef – Ei = (Wother)i6f
Wother includes all other forces that you did not account for in
potential energies. Thus it might include some conservative
forces you did not care to incorporate into potential energy, but
always includes all non-conservative forces.
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Ans: Energy
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Want V.
Example: A block of mass M is on a level frictionless surface
and is used to compress a spring, of force constant k, by a
distance L. The block is released from rest while a constant
wind blows against it with force magnitude B, down by angle
θ from the horizontal. What is its speed V after it has moved
distance 3/2 L?
3 forces:
PF , PN, PT
g
ΔE = Ef – Ei = (Wother)i6f
[ Kf + Ugf ] S [Ki + Ugi ]
= WN + WT =
∫
Nid
0
+P
T CD
P
[ &frac12;mvf2 + mgyf ] S [ &frac12;mvi2 + mgyi ] = |T
P| DT (…TDcosθ !)
Note: DT = +(L+d) (how much of P
D is in the direction of P
T)
[ &frac12;mV2 + mg(S h) ] S [ &frac12;m(0)2 + mg(0) ] = T [+(L+d)]
V2 = 2gh + 2 T (L+d) /m
(note: θ is a distractor)
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Example: A block of mass M is on a
level frictionless surface and is used
to compress a spring, of force
constant k, by a distance L. The
block is released from rest while a
constant wind blows against it with
force magnitude B, down by angle θ
from the horizontal. What is its speed
V after it has moved distance 3/2 L?
NON-MECHANICAL ENERGY
ΔE = Ef S Ei = (Wother)i6f
where E = K + U = mechanical energy
Ef S Ei = (Wother)i6f
PF
g
&amp;P
F S conservative Y Ug &amp; US, but treat P
N and P
B as “other”
[&frac12;MVf2 + Mgyf + &frac12;ksf2] – [&frac12;MVi2 +Mgyi + &frac12;ksi2] = WN + WB
&frac12;MV2 +
MgY
+ &frac12;k(0)2 S [0 +
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MgY + &frac12;k( S L)2] = P
N •D
P + PB •DP
&frac12;MV2 S &frac12; k L2 = 0 + BD D
&frac12;MV2 S &frac12; k L2 = ( S B cosθ) (3/2 L)
We can change this into an even more powerful form
by noting that (Wother)i6f is an energy term, and is a
means of generating non-mechanical energy.
A wonderful example is friction. If you rub your
hands together, you will note two things:
C you can hear it (sound energy)
C they warm up (heat energy)
and then solve for V.
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CONSERVATION OF TOTAL ENERGY
We can just lump the non-mechanical energies (e.g.,
heat energy, internal chemical energies, etc.) into a
generic “other energy” term gother (or equivalently,
internal ginternal) that a system has. We define a quantity
called TOTAL ENERGY &otilde; to be the sum of MECHANICAL
ENERGY E and OTHER ENERGY g:
Example: A small block is on an undulating track that is
frictionless except for two rough patches. It is initially
pressed against a pivoted spring of force constant k that has
been compressed distance d. It is released from rest at
height H above the bottom of the track. The block comes to
rest in the midst of the second rough patch at height h = aH.
What is change in internal energy Δg (mainly heating)?
&otilde;= E+g = K+U +g
The total energy of a “closed system” is conserved,
giving THE LAW:
&otilde;i = &otilde;f
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&otilde;f = &otilde;i
Ef + gf =
Ei + gi
gf S gi = Ei S Ef
Δ g = Ei S Ef
(NOTE : Δ g = S ΔE = S (Wfric)i6f )
Δ g = [&frac12; m'
v i2 + m g yi + &frac12; k si2 ] – [&frac12; m'
v f2 + m g yf + &frac12; k '
s f2 ]
= [ m g (+ H) + &frac12; k(S d)2 ] S [ m g (+aH) ]
Y
Δ g = + b m g H + &frac12; k d2 = S (Wfric)i6f
Increase in internal energy equals the loss of mechanical
energy due to the work of friction in a closed system.
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