Download Phys 207 E = K + U is constant!!!

Phys 207
Announcements
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Exam 2 this Friday, April 8th
Today’s Agenda
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Review:
ÍConservation of “total mechanical energy”
ÍNon-conservative forces (friction)
General work/energy theorem
Problems using work/energy theorem
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Conservation of Energy
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If only conservative forces are present, the total kinetic plus
potential energy of a system is conserved, i.e. the total
“mechanical energy” is conserved.
Í(note: E=Emechanical throughout this discussion)
E=K+U
∆E = ∆K + ∆U
⇒ using ∆K = W
= W + ∆U
= W + (-W) = 0 ⇒ using ∆U = -W
E = K + U is constant!!!
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Both K and U can change, but E = K + U remains constant.
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Page 1
Problem: Hotwheel
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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.
Í Find v1 and v2.
v2
d
h
v1
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Problem: Hotwheel...
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K+U energy is conserved, so ∆E = 0
∆K = - ∆U
Moving down a distance d, ∆U = -mgd, ∆K = 1/2mv12
Solving for the speed:
v1 = 2 gd
d
h
v1
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Page 2
Problem: Hotwheel...
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At the end, we are a distance d - h below our starting point.
∆U = -mg(d - h), ∆K = 1/2mv22
Solving for the speed:
v 2 = 2 g (d − h )
d-h
v2
d
h
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Non-conservative Forces: Friction
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Since the force is constant in magnitude and opposite in direction to
the displacement, the work done in pushing the box through an
arbitrary path of length L is just
Wf = -µmgL.
Clearly, the work done depends on the path taken.
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Wpath 2 > Wpath 1
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B
path 1
path 2
A
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When non-conservative forces act then energy can be dissipated into
other modes (thermal,sound)
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Page 3
Generalized Work/Energy Theorem:
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Suppose FNET = FC + FNC (sum of conservative and nonconservative forces).
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The total work done is: WNET = WC + WNC
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The Work/Kinetic Energy theorem says that: WNET = ∆K.
Í WNET = WC + WNC = ∆K
Í
WNC = ∆K - WC
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But WC = -∆U
So
WNC = ∆K + ∆U = ∆Emechanical
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Generalized Work/Energy Theorem:
WNC = ∆K + ∆U = ∆Emechanical
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The change in kinetic+potential energy of a system is equal
to the work done on it by non-conservative forces.
Emechanical =K+U of system not conserved!
ÍIf all the forces are conservative, we know that K+U
energy is conserved: ∆K + ∆U = ∆Emechanical = 0 which says
that WNC = 0, which makes sense.
ÍIf some non-conservative force (like friction, a “push” or a
“pull”) does work, K+U energy will not be conserved and
WNC = ∆E, which also makes sense.
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Page 4
Problem: Block Sliding with Friction
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A block slides down a frictionless ramp. Suppose the horizontal
(bottom) portion of the track is rough, such that the coefficient of
kinetic friction between the block and the track is µk.
Í How far, x, does the block go along the bottom portion of the
track before stopping?
d
µk
x
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Problem: Block Sliding with Friction...
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Using WNC = ∆K + ∆U
As before, ∆U = -mgd
WNC = work done by friction = -µkmgx.
∆K = 0 since the block starts out and ends up at rest.
WNC = ∆U
-µkmgx = -mgd
x = d / µk
d
µk
x
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Page 5
More Example Problems
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Problems using work/energy theorem
ÍSpring shot
ÍEscape velocity
ÍLoop the loop
ÍVertical springs
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Note: Lecture notes will have additional, related
examples worked out.
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Problem: Spring Shot
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A sling shot is made from a pair of springs each having spring
constant k. The initial length of each spring is x0. A puck of mass
m is placed at the point connecting the two springs and pulled
back so that the length of each spring is x1. The puck is released.
What is its speed v after leaving the springs? (The relaxed length
of each spring is xr).
xr
x1
x0
m
m
m
v
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Page 6
Problem: Spring Shot
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Only conservative forces are at work, so K+U energy is
conserved. EI = EF
∆K = -∆Us
(
) (
1
2
2
2
2
∆U s = 2 ⋅ k ( x0 − x r ) − ( x1 − x r ) = k ( x0 − x r ) − ( x1 − x r )
2
)
x1
x0
m
m
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Problem: Spring Shot
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Only conservative forces are at work, so K+U energy is
∆K = -∆Us
conserved. EI = EF
∆K =
1
mv 2
2
m
m at rest
v
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Page 7
Problem: Spring Shot
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Only conservative forces are at work, so K+U energy is
conserved. EI = EF
∆K = -∆Us
(
1
2
2
mv 2 = −k ( x0 − x r ) − ( x1 − x r )
2
m
)
m
v
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Problem: How High?
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A projectile of mass m is launched straight up from the surface of
the earth with initial speed v0. What is the maximum distance
from the center of the earth RMAX it reaches before falling back
down.
RMAX
RE
m
v0
M
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Page 8
Problem: How High...
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All forces are conservative:
ÍWNC = 0
Í∆K = -∆U
RMAX
RE
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And we know:
1
∆K = − mv 02
2
⎛ 1
1
− ∆U = −GMm⎜
−
⎜R
R
⎝ E
MAX
m
⎞
⎟
⎟
⎠
v0
hMAX
⎛ 1
1
1
mv 02 = GMm ⎜⎜
−
2
⎝ RE RMAX
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⎞
⎟
⎟
⎠
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Problem: How High...
⎛ 1
1
1
−
mv 02 = GMm⎜⎜
2
⎝ RE RMAX
⎛ 1
1
−
v 02 = 2GM ⎜
⎜R
⎝ E RMAX
⎞
⎟
⎟
⎠
RMAX
⎞
⎟
⎟
⎠
⎛ GM ⎞ ⎛
R
= 2 ⎜ 2 ⎟RE ⎜1 − E
⎜R ⎟ ⎜
RMAX
⎝ E ⎠ ⎝
⎛
R
= 2 gRE ⎜1 − E
⎜
RMAX
⎝
v 02
R
=1 − E
2 gRE
RMAX
⎞
⎟
⎟
⎠
RE
m
v0
hMAX
⎞
⎟
⎟
⎠
M
R MAX =
RE
v 02
1−
2 gRE
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Page 9
Escape Velocity
R MAX =
RE
v 02
1−
2 gRE
If we want the projectile to escape to infinity we need to make the
denominator in the above equation zero:
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1−
v 02
=0
2 gR E
v 02
=1
2 gRE
v 0 = 2 gRE
We call this value of v0 the escape velocity, vesc
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Escape Velocity
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GM
we find the escape velocity from
RE2
GM p
a planet of mass Mp and radius Rp to be:
v esc = 2
Rp
(where G = 6.67 x 10-11 m3 kg-1 s-2).
Remembering that g =
Rp(m)
Mp(kg)
gp(m/s2)
vesc(m/s)
Earth
6.378x106
5.976x1024
9.81
11.2x103
Moon
1.737x106
7.349x1022
1.62
2.38x103
Jupiter
7.149x107
1.900x1027
24.8
59.5x103
Sun
6.950x108
1.989x1030
275
618.x103
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Page 10
Lecture 18, Act 1
Escape Velocity
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Two identical spaceships are awaiting launch on two planets with the
same mass. Planet 1 is stationary, while Planet 2 is rotating with an
angular velocity ω.
ÍWhich spaceship needs more fuel to escape to infinity?
(a) 1
(b) 2
(c) same
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(1)
ω
(2)
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Lecture 18, Act 1
Solution
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Both spaceships require the same escape velocity to reach
infinity.
Thus, they require the same kinetic energy.
Both initially have the same potential energy.
Spaceship 2 already has some kinetic energy due to its
rotational motion, so it requires less work (i.e. less fuel).
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Page 11
Lecture 18, Act 1
Algebraic Solution
WNC = ∆K + ∆U = ∆E
For spaceship 1:
W1 = (Kf - K0) + (Uf - U0)
K0 = 0 , U0 = 0
W1 = Kf + Uf
For spaceship 2:
W2 = (Kf - K0) + (Uf - U0) =
K0 =
1
m(ωr)2, U0 = 0
2
W2 = Kf + Uf -
1
m(ωr)2
2
W1 > W2 So spaceship 1 will need more fuel.
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v = ωr
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Lecture 18, Act 1
Aside
This is one of the reasons why all of the world’s spaceports
are located as close to the equator as possible.
r1
r2 > r1
r2
K2 =
1
1
m(ωr2)2 > K1 = m(ωr1)2
2
2
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Page 12
Problem: Space Spring
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A low budget space program
decides to launch a 10,000 kg
spaceship into space using a big
spring. If the spaceship is to reach
a height RE above the surface of
the Earth, what distance d must
the launching spring be
compressed if it has a spring
constant of 108 N/m.
d
k
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Problem: Space Spring...
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Since gravity is a conservative force, energy is conserved. Since
K = 0 both initially and at the maximum height (v = 0) we know:
Ubefore = Uafter
(US + UG )before = (UG )after
⎛ 1 2 GMm ⎞ ⎛ GMm ⎞
⎟ = ⎜−
⎟
⎜ kd −
⎜
RE ⎟⎠ ⎝⎜ 2 RE ⎟⎠
⎝2
1
GMm 2 GMm GMm
kd 2 = −
+
=
2
2 RE
2 RE
2 RE
d =
GMm
kRE
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Page 13
Problem: Space Spring
So we find
d =
GMm
kRE
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For the numbers given, d = 79.1 m
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But don’t get too happy...
F = kd = ma
a = kd/m
a = 79.1 x 106 m/s2
a = 791000 g
unhappy astronaut!
d
a
k
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Problem: Loop the loop
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A mass m starts at rest on a frictionless track a distance H
above the floor. It slides down to the level of the floor
where it encounters a loop of radius R. What is H if the
mass just barely makes it around the loop without losing
contact with the track.
H
R
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Page 14
Problem: Loop the loop
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Draw a FBD of the mass at
the top of the loop:
FTOT = -(mg+N) j
ma = -mv2/R j
If it “just” makes it, N = 0.
Ímg = mv2/R
N
v
mg
v = Rg
v
j
H
R
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v = Rg
Problem: Loop the loop
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Now notice that K+U energy is conserved. ∆K = -∆U.
∆U = -mg(h) = -mg(H-2R)
∆K = 1/2 mv2 = 1/2 mRg
H=
mg(H-2R) = 1/2 mRg
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R
2
h = H - 2R
v
H
R
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Page 15
Lecture 18, Act 2
Energy Conservation
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A mass starts at rest on a frictionless track a distance H above the
floor. It slides down to the level of the floor where it encounters a
loop of radius R. What is H if the normal force on the block by the
track at the top of the loop is equal to the weight of the block ?
(a) 3R
(b) 3.5R
(c) 4R
H
R
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Lecture 18, Act 2
Solution
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Draw a FBD of the mass at
the top of the loop:
FNET = -(mg+N) j
ma = -mv2/R j
In this case, N = mg.
Í2mg = mv2/R
N
v
mg
v 2 = 2Rg
v
j
H
R
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v 2 = 2Rg
Lecture 18, Act 2
Solution
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Use the fact that K+U energy is conserved: ∆K = -∆U.
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∆U = -mg(h) = -mg(H - 2R), ∆K = 1/2 mv2 = mRg
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mg(H - 2R) = mRg
H = 3R
h = H - 2R
v
H
R
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Vertical Springs
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A spring is hung vertically. Its
relaxed position is at y = 0 (a). When
a mass m is hung from its end, the
new equilibrium position is ye (b).
(b)
(a)
j
k
Recall that the force of a spring
is Fs = -kx. In case (b)
Fs = mg and x = ye:
-kye - mg = 0
y=0
(ye < 0)
m
mg = -kye
mg
y = ye
-kye
(ok since ye is a negative number)
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Page 17
Vertical Springs
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U =
(b)
(a)
The potential energy of the springmass system is:
j
1
ky 2 + mgy + C
2
k
but mg = -kye
y=0
1
U = ky 2 − ky e y + C
2
choose C to make U=0 at y = ye:
0=
1
2
2
ky e − ky e + C
2
C=
m
y = ye
1
2
ky e
2
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Vertical Springs
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(b)
(a)
So:
j
1
1
U = ky 2 - ky e y + ky e 2
2
2
1
2
2
= k y + y e - 2y e y
2
(
k
)
y=0
which can be written:
m
1
2
U = k (y − y e )
2
y = ye
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Page 18