Physics 2211A Todays Agenda

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Physics 2211ATodays Agenda

• Review
• Work done by variable force in 3-D
• Conservative forces potential energy
• Conservation of total mechanical energy
• Examples

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• A car manufacturer claims that his car can
  accelerate from rest to 100 km/hr in 8.0 s. The
  cars mass is 900 kg. Determine the average
  power developed by the cars engine.
• (1) 87 kW (2) 0.004 kW
• (3) 43 kW (4) 560 kW
• (5) 34 kW

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Work by variable force in 3-D

• Work dWF of a force F acting
• through an infinitesimal
• displacement ?r is
• dW F.?r
• The work of a big displacement through a variable
  force will be the integral of a set of
  infinitesimal displacements
• WTOT F.?r

ò
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Conservative Forces

• We have seen that the work done by gravity does
  not depend on the path taken.

m
R2
R1
M
m
h
Wg -mgh
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Conservative Forces

• In general, if the work done does not depend on
  the path taken, the force involved is said to be
  conservative.
• Gravity near the Earths surface
• A spring produces a conservative force

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Conservative Forces

• We have seen that the work done by a conservative
  force does not depend on the path taken.

W2
W1 W2
W1
Therefore the work done in a closed path is 0.
W2
WNET W1 - W2 0
W1
The work done can be reclaimed.
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• The pictures below show force vectors at
  different points in space for two forces. Which
  one is conservative ?

(a) 1 (b) 2 (c)
both
y
y
(1)
(2)
x
x
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• Consider the work done by force when moving along
  different paths in each case

WA WB
WA gt WB
(1)
(2)
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• In fact, you could make money on type (2) if it
  ever existed
• Work done by this force in a round trip is gt 0!
• Free kinetic energy!!

WNET 10 J DK
W 15 J
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Potential Energy

• For any conservative force F we can define a
  potential energy function U in the following
  way
• The work done by a conservative force is equal
  and opposite to the change in the potential
  energy function.
• This can be written as

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Gravitational Potential Energy

• We have seen that the work done by gravity near
  the Earths surface when an object of mass m is
  lifted a distance ?y is Wg -mg ?y
• The change in potential energy of this object is
  therefore ?U -Wg mg ?y

m
j
?y
Wg -mg ?y
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Gravitational Potential Energy

• So we see that the change in U near the Earths
  surface is ?U -Wg mg ?y mg(y2 -y1).
• So U mg y U0 where U0 is an arbitrary
  constant.
• Having an arbitrary constant U0 is equivalent to
  saying that we can choose the y location where U
  0 to be anywhere we want to.

m
j
y2
Wg -mg ?y
y1
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Potential Energy Recap

• For any conservative force we can define a
  potential energy function U such that
• The potential energy function U is always defined
  onlyup to an additive constant.
• You can choose the location where U 0 to be
  anywhere convenient.

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Conservative Forces Potential Energies (stuff
you should know)
Work W(1-2)
Change in P.E ?U U2 - U1
P.E. function U
Force F

Fg -mg j
-mg(y2-y1)
mg(y2-y1)
mgy C
Fs -kx
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• All springs and masses are identical. (Gravity
  acts down).
• Which of the systems below has the most potential
  energy stored in its spring(s), relative to the
  relaxed position?

(1) 1 (2) 2 (3) same
(1)
(2)
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• The displacement of (1) from equilibrium will be
  half of that of (2) (each spring exerts half of
  the force needed to balance mg)

0
d
2d
(1)
(2)
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0
d
2d
(1)
(2)
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Conservation of Energy

• If only conservative forces are present, the
  total kinetic plus potential energy of a system
  is conserved.
  E K U is constant!!!
• Both K and U can change, but E K U remains
  constant.

E K U ?E ?K ?U W ?U W
(-W) 0
using ?K W using ?U -W
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Example The simple pendulum

• Suppose we release a mass m from rest a distance
  h1 above its lowest possible point.
• What is the maximum speed of the mass and
  wheredoes this happen?
• To what height h2 does it rise on the other side?

m
h1
h2
v
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Example The simple pendulum

• Kineticpotential energy is conserved since
  gravity is a conservative force (E K U is
  constant)
• Choose y 0 at the bottom of the swing, and U
  0 at y 0 (arbitrary choice)E 1/2mv2 mgy

y
h1
h2
y 0
v
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Example The simple pendulum

• E 1/2mv2 mgy.
• Initially, y h1 and v 0, so E mgh1.
• Since E mgh1 initially, E mgh1 always since
  energy is conserved.

y
y 0
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Example The simple pendulum

• 1/2mv2 will be maximum at the bottom of the
  swing.
• So at y 0 1/2mv2 mgh1 v2
  2gh1

y
y h1
h1
y 0
v
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Example The simple pendulum

• Since E mgh1 1/2mv2 mgy it is clear that
  the maximum height on the other side will be at y
  h1 h2 and v 0.
• The ball returns to its original height.

y
y h1 h2
y 0
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Example The simple pendulum

• The ball will oscillate back and forth. The
  limits on its height and speed are a consequence
  of the sharing of energy between K and U. E
  1/2mv2 mgy K U constant.

y
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Example The simple pendulum

• We can also solve this by choosing y 0 to be at
  the original position of the mass, and U 0 at y
  0.E 1/2mv2 mgy.

y
y 0
h1
h2
v
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Example The simple pendulum

• E 1/2mv2 mgy.
• Initially, y 0 and v 0, so E 0.
• Since E 0 initially, E 0 always since energy
  is conserved.

y
y 0
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Example The simple pendulum

• 1/2mv2 will be maximum at the bottom of the
  swing.
• So at y -h1 1/2mv2 mgh1
  v2 2gh1

y
Same as before!
y 0
h1
y -h1
v
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Example The simple pendulum

• Since 1/2mv2 - mgh 0 it is clear that the
  maximum height on the other side will be at y 0
  and v 0.
• The ball returns to its original height.

y
y 0
Same as before!
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Example Airtrack Glider

• A glider of mass M is initially at rest on a
  horizontal frictionless track. A mass m is
  attached to it with a massless string hung over a
  massless pulley as shown. What is the speed v of
  M after m has fallen a distance d ?

v
M
m
d
v
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Example Airtrack Glider

• Kineticpotential energy is conserved since all
  forces are conservative.
• Choose initial configuration to have U0.?K
  -?U

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Problem Hotwheel

• A toy car slides on the frictionless track shown
  below. It starts at rest, drops a distance d,
  moves horizontally at speed v1, rises a distance
  h, and ends up moving horizontally with speed v2.
• Find v1 and v2.

v2
d
h
v1
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Problem Hotwheel...

• KU energy is conserved, so ?E 0 ?K -
  ?U
• Moving down a distance d, ?U -mgd, ?K
  1/2mv12
• Solving for the speed

d
h
v1
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Problem Hotwheel...

• At the end, we are a distance d - h below our
  starting point.
• ?U -mg(d - h), ?K 1/2mv22
• Solving for the speed

v2
d - h
d
h
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Recap of todays lecture

• Work done by variable force in 3-D
• Conservative Forces Potential energy
• Conservation of Total Mechanical Energy
• Examples pendulum, airtrack, Hotwheel car