Physics 2211: Lecture 17 Todays Agenda

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Physics 2211 Lecture 17Todays Agenda

• Work Energy
• Review
• Work done by variable forces
• Springs
• Power

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Review Constant Force
Work, W, of a constant force F acting through
a displacement ?r is W F? ?r F ?r cos(?)
Fr ?r
F
?r
?
Fr
displacement
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Work done by Variable Force (1D)

• When the force was constant, we wrote W F ?x
• area under F vs. x plot
• For variable force, we find the areaby
  integrating
• dW F(x) dx.

F
Wg
x
?x
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Work/Kinetic Energy Theorem for a Variable Force
dv
F
dx
F
dx
dv
dv
dv
v
(chain rule)
dx


dx
dt
dx
dt
dv
dx
v
dx
v dv
v22
v12
v22
v12
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• A force A3i -4j 5k (N) acts on an object as it
  undergoes a displacement of 1i2j3k. What is
  the work done by the force?
• (1) 10 J (2) 20 J
• (2) 60 J (3) 26 J
• (5) 3.6 J

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11331

• The graph F (in N) vs x (in m) is shown for the
  net force F acting on a 0.25 kg object that moves
  only along the x axis. If the object has a
  velocity of 7 m/s at x0, what is the maximum
  kinetic energy between x0 m and x9 m?
• (1) 6.1 J (2) 14 J
• (3) 9.1 J (4) 10 J
• (5) 8.1 J

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F
X
9
5
6
7
8
4
1
2
3
-2
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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?

(1) 1 (2) 2 (3) 3
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• First, draw all the forces in the system

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• Recall that W F? ?r so only forces that have
  a component along the direction of the
  displacement are doing work.

FS
a
N
mg
The answer is (b) 2.
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Springs

• Hookes Law The force exerted by a spring is
  proportional to the distance the spring is
  stretched or compressed from its relaxed
  position.
• FX -k x Where x is the displacement from
  the relaxed position and k is the constant
  of proportionality.

relaxed position
FX 0
x
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Springs...

• Hookes Law The force exerted by a spring is
  proportional to the distance the spring is
  stretched or compressed from its relaxed
  position.
• FX -k x Where x is the displacement from
  the relaxed position and k is the constant
  of proportionality.

relaxed position
FX -kx gt 0
x
x ? 0
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Springs...

• Hookes Law The force exerted by a spring is
  proportional to the distance the spring is
  stretched or compressed from its relaxed
  position.
• FX -k x Where x is the displacement from the
  relaxed position and k is the constant of
  proportionality.

relaxed position
FX - kx lt 0
x
x gt 0
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Scales

• Springs can be calibrated to tell us the applied
  force.
• We can calibrate scales to read Newtons, or...
• Fishing scales usually read weight in kg or lbs.

1 lb 4.45 N
0
2
4
6
8
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1-D Variable Force Example Spring

• For a spring we know that Fx -kx.

F(x)
x2
x1
x
relaxed position
-kx
F - k x1
F - k x2
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Spring...

• 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)
x2
x1
x
Ws
relaxed position
-kx
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Spring...

• 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)
x2
x1
x
Ws
-kx
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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.
• If the initial speed of the box were doubled and
  its mass were halved, how far x2 would the spring
  compress ?

(a) (b)
(c)
x
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• Again, use the fact that WNET DK.

In this case, WNET WSPRING -1/2 kx2 and
?K -1/2 mv2
In the case of x1
so kx2 mv2
x1
v1
m1
m1
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So if v2 2v1 and m2 m1/2
x2
v2
m2
m2
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Problem Spring pulls on mass.

• 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
stretched position (at rest)
m
d
after release
m
v
back at relaxed position
m
vr
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Problem Spring pulls on mass.

• First find the net work done on the mass during
  the motion from x d to x 0 (only due to the
  spring)

stretched position (at rest)
m
d
relaxed position
m
i
vr
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Problem Spring pulls on mass.

• Now find the change in kinetic energy of the mass

stretched position (at rest)
m
d
relaxed position
m
i
vr
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Problem Spring pulls on mass.

• Now use work kinetic-energy theorem Wnet WS
  ?K.

stretched position (at rest)
m
d
relaxed position
m
i
vr
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Problem Spring pulls on mass.

• Now suppose there is a coefficient of friction ?
  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 - ?mg d

?r
stretched position (at rest)
m
d
relaxed position
f ?mg
m
i
vr
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Problem Spring pulls on mass.

• Again use Wnet WS Wf ?K Wf -?mg d

?r
stretched position (at rest)
m
d
relaxed position
f ?mg
m
i
vr
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Power

• We have seen that W F.?r
• This does not depend on time!

F
?r

• Units of power J/sec N-m/sec Watts

v
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Power

• A 2000 kg trolley is pulled up a 30 degree hill
  at 20 mi/hrby a winch at the top of thehill.
  How much power is thewinch providing?
• The power is P F.v T.v
• Since the trolley is not accelerating, the net
  force on it must be zero. In the x direction
• T - mg sin ? 0
• T mg sin ?

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Power

• P T.v Tv since T is parallel to v
• So P mgv sin ?
• v 20 mi/hr 8.94 m/sg 9.81 m/s2m 2000
  kgsin ? sin(30o) 0.5
• and P (2000 kg)(9.81 m/s2)(8.94 m/s)(0.5)
  87,700 W

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Recap of todays lecture

• Work done by Springs
• Power