Download Lecture 15 - UD Physics

Phys 207
Announcements
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Hwk 6 is posted online; submission deadline = April 4
Exam 2 on Friday, April 8th
Today’s Agenda
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Review of Work & Energy
(Chapter 7)
Work of multiple constant forces
Work done by gravity near the Earth’s surface
Examples:
Ípendulum, inclined plane, free fall
Work done by variable force
ÍSpring
Problem involving spring & friction
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Review: Constant Force
Work, W, of a constant force F
acting through a displacement ∆r
is:
W = Fi ∆r = F ∆r cos(θ) = Fr ∆r
F
θ
dis
Fr
∆r
t
en
m
ce
pla
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Page 1
Review: Work/Kinetic Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
WF = ∆K = 1/2mv22 - 1/2mv12
v2
v1
F
WF = F∆x
m
∆x
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A simple application:
Work done by gravity on a falling object
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What is the speed of an object after falling a distance H,
assuming it starts at rest?
Wg = Fi ∆r = mg ∆r cos(0) = mgH
v0 = 0
Wg = mgH
H
∆r
mg
j
Work/Kinetic Energy Theorem:
Wg = mgH = 1/2mv2
v = 2 gH
v
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Page 2
What about multiple forces?
Suppose FNET = F1 + F2 and the
displacement is ∆r.
The work done by each force is:
W1 = F1i ∆r
F1
W2 = F2 i ∆r
FNET
WTOT = W1 + W2
= F1i ∆r + F2i ∆r
= (F1 + F2 )i ∆r
WTOT = FTOTi ∆r
∆r
F2
It’s the total force that matters!!
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Comments:
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Time interval not relevant
ÍRun up the stairs quickly or slowly...same W
Since W = Fi ∆r
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No work is done if:
ÍF = 0
or
Í∆r = 0
or
Íθ = 90o
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Page 3
Comments...
W = Fi ∆r
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No work done if θ = 90o.
T
ÍNo work done by T.
v
v
N
No work done by N.
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Lecture 15, Act 1
Work & Energy
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An inclined plane is accelerating with constant acceleration a.
A box resting on the plane is held in place by static friction.
How many forces are doing work on the block?
a
(a) 1
(b) 2
(c) 3
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Page 4
Lecture 15, Act 1
Solution
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First, draw all the forces in the system:
FS
a
mg
N
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Lecture 15, Act 1
Solution
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Recall that W = Fi ∆r so only forces that have a
component along the direction of the displacement are
doing work.
FS
a
mg
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N
The answer is (b) 2.
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Page 5
Work done by gravity:
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Wg = Fi ∆r = mg ∆r cos θ
= -mg ∆y
m
mg
Wg = -mg ∆y
∆r θ
j
−∆y
Depends only on ∆y !
m
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Work done by gravity...
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W NET = W1 + W2 + . . .+ Wn
= Fi ∆r 1+ Fi ∆r2 + . . . + Fi ∆rn
= Fi (∆r1 + ∆r 2+ . . .+ ∆rn)
= Fi ∆r
= F ∆y
m
∆y
∆r3
Wg = -mg ∆y
Depends only on ∆y,
not on path taken!
∆r
∆r1
∆r2
mg
j
∆rn
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Page 6
Lecture 15, Act 2
Falling Objects
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Three objects of mass m begin at height h with velocity 0. One
falls straight down, one slides down a frictionless inclined
plane, and one swings on the end of a pendulum. What is the
relationship between their velocities when they have fallen to
height 0?
v=0
v=0
v=0
H
vf
vp
vi
Free Fall
Frictionless incline
(a) Vf > Vi > Vp
(b) Vf > Vp > Vi
Pendulum
(c) Vf = Vp = Vi
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Lecture 15, Act 2
Solution
v=0
v=0
v=0
H
vf
Free Fall
vp
vi
Frictionless incline
Pendulum
Only gravity will do work: Wg = mgH = 1/2 mv22 - 1/2 mv12 = 1/2 mv22
vf = v i = v p = 2gH
does not depend on path !!
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Page 7
Lifting a book with your hand:
What is the total work done on the book??
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First calculate the work done by gravity:
Wg = mgi ∆r = -mg ∆r
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FHAND
∆r
Now find the work done by
the hand:
v = const
a=0
WHAND = FHANDi ∆r = FHAND ∆r
mg
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Example: Lifting a book...
Wg
= -mg ∆r
WHAND = FHAND ∆r
∆r
WNET
FHAND
v = const
a=0
= WHAND + Wg
= FHAND ∆r - mg ∆r
= (FHAND - mg) ∆r
mg
= 0 since ∆K = 0 (v = const)
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So WTOT = 0!!
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Page 8
Example: Lifting a book...
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Work/Kinetic Energy Theorem says: W = ∆K
{Net Work done on object} = {change in kinetic energy of object}
In this case, v is constant so ∆K = 0
and so W must be 0, as we found.
∆r
FHAND
v = const
a=0
mg
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Work done by Variable Force: (1D)
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When the force was constant, we
wrote W = F ∆x
Íarea under F vs. x plot:
F
Wg
x
∆x
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For variable force, we find the area
by integrating:
ÍdW = F(x) dx.
F(x)
x2
W = ∫ F ( x )dx
x1
x1
dx
x2
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Page 9
Work/Kinetic Energy Theorem for a
Variable Force
x2
F = ma = m dv
dt
W = ∫ F dx
x1
x
= m ∫ dv dx
x dt
dv dx dv
dv
=
= v dx (chain rule)
dt
dt dx
2
1
v2
=m∫v
v1
dv
dx
dx
v2
= m ∫ v dv
v1
1
1
1
= m (v22 −v12 ) = m v22 − m v12 = ∆KE
2
2
2
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1-D Variable Force Example: Spring
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For a spring we know that Fx = -kx.
F(x)
x1
x2
x
relaxed position
-kx
F = - k x1
F = - k x2
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Page 10
Spring...
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The work done by the spring Ws during a displacement
from x1 to x2 is the area under the F(x) vs x plot between
x1 and x2.
F(x)
x1
x2
x
Ws
relaxed position
-kx
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Spring...
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The work done by the spring Ws during a displacement
from x1 to x2 is the area under the F(x) vs x plot between
x1 and x2.
x2
F(x)
x1
Ws = ∫ F ( x )dx
x1
x2
x2
x
Ws
= ∫ ( −kx )dx
x1
=−
-kx
Ws = −
1 2
kx
2
x2
x1
1
k (x22 − x12 )
2
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Page 11
Lecture 15, Act 3
Work & Energy
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A box sliding on a horizontal frictionless surface runs into a
fixed spring, compressing it a distance x1 from its relaxed
position while momentarily coming to rest.
ÍIf the initial speed of the box were doubled and its mass
were halved, how far x2 would the spring compress ?
(a)
x2 = x1
(b) x2 = 2 x1
(c)
x2 = 2 x1
x
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Lecture 15, Act 3
Solution
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Again, use the fact that WNET = ∆K.
WNET = WSPRING = -1/2 kx2
∆K = -1/2 mv2
In this case,
and
so kx2 = mv2
In the case of x1
x1 = v1
m1
k
x1
v1
m1
m1
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Page 12
x =v
m
k
Lecture 15, Act 3
Solution
So if v2 = 2v1 and m2 = m1/2
x2 = 2v1
m1 2
k
= v1
2m1
k
x2 = 2x1
x2
v2
m2
m2
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Problem: Spring pulls on mass.
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A spring (constant k) is stretched a distance d, and a mass m
is hooked to its end. The mass is released (from rest). What
is the speed of the mass when it returns to the relaxed
position if it slides without friction?
m
relaxed position
m stretched position (at rest)
d
m
after release
v
back at relaxed position
m
vr
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Page 13
Problem: Spring pulls on mass.
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First find the net work done on the mass during the motion
from x = d to x = 0 (only due to the spring):
Ws = −
1
1
1
k (x 22 − x12 ) = − k (0 2 − d 2 ) = kd 2
2
2
2
m stretched position (at rest)
d
relaxed position
m
i
vr
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Problem: Spring pulls on mass.
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Now find the change in kinetic energy of the mass:
∆K =
1
1
1
mv 22 − mv12 = mv r2
2
2
2
m stretched position (at rest)
d
relaxed position
m
i
vr
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Page 14
Problem: Spring pulls on mass.
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Now use work kinetic-energy theorem: Wnet = WS = ∆K.
1
kd 2 =
2
1
mv r 2
2
vr = d
k
m
m stretched position (at rest)
d
relaxed position
m
i
vr
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Problem: Spring pulls on mass.
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Now suppose there is a coefficient of friction µ between the
block and the floor
The total work done on the block is now the sum of the work
done by the spring WS (same as before) and the work done by
friction Wf.
.
Wf = f ∆r = - µmg d
∆r
m stretched position (at rest)
d
m
f = µmg
relaxed position
i
vr
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Page 15
Problem: Spring pulls on mass.
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Again use Wnet = WS + Wf = ∆K
1
WS = kd 2
Wf = -µmg d
2
∆K =
1 2
1
2
kd − µmgd = mv r
2
2
vr =
1
2
mv r
2
k 2
d − 2 µgd
m
∆r
m stretched position (at rest)
d
m
f = µmg
relaxed position
i
vr
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