Download Lecture 8.2

Welcome back to Physics 215
Today’s agenda:
• More gravitational potential energy
• Potential energy of a spring
• Work-kinetic energy theorem
Physics 215 – Fall 2014
Lecture 08-2
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Current homework assignment
• HW7:
– Knight Textbook Ch.9: 54, 72
– Ch.10: 48, 68, 76
– Ch.11: 50, 64
– Due Wednesday, Oct. 22nd in recitation
Physics 215 – Fall 2014
Lecture 08-2
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Gravitational Potential Energy
For an object of mass m near the surface of the earth:
Ug = mgh
• h is height above arbitrary reference line
• Measured in Joules -- J (like kinetic energy)
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Total energy for object moving
under gravity
E = Ug + K = constant
* E is called the (mechanical) energy
* It is conserved:
(½) mv2 + mgh = constant
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A ball of mass m=7 kg attached to a massless string of length
R=3 m is released from the position shown in the figure below.
(a) Find magnitude of velocity of the ball at the lowest point on its
path. (b) Find the tension in the string at that point.
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Stopped-pendulum demo
• Pendulum swings to same height on other
side of vertical
• What if pendulum string is impeded ~1/2way along its length? Will height on other
side of vertical be:
1. Greater than original height
2. Same as original height
3. Less than original height?
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A block is released from rest on a frictionless
incline. The block travels to the bottom of the
left incline and then moves up the right incline
which is steeper than the left side.
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Springs -- Elastic potential energy
frictionless table
Force F = -kx (Hooke’s law)
F
Area of triangle lying under
straight line graph of F vs. x
= (1/2)(+/-x)(-/+kx)
Us =
(1/2)kx2
Physics 215 – Fall 2014
x
F = -kx
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(Horizontal) Spring
frictionless
table
• x = displacement from relaxed state of spring
• Elastic potential energy stored in spring: Us = (1/2)kx2
(1/2)kx2 + (1/2)mv2 = constant
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A 0.5 kg mass is attached to a spring on a horizontal
frictionless table. The mass is pulled to stretch the spring
5.0 cm and is released from rest. When the mass crosses
the point at which the spring is not stretched, x = 0, its
speed is 20 cm/s. If the experiment is repeated with a
10.0 cm initial stretch, what speed will the mass have
when it crosses x = 0 ?
1. 40 cm/s
2. 0 cm/s
3. 20 cm/s
4. 10 cm/s
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Mass hanging on spring
• Now oscillations are about
equilibrium point of spring + mass
• Otherwise, motion is same as
horizontal mass + spring on
frictionless table
(1/2)mv2 = (1/2)ka2 - (1/2)kz2
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Work, Energy
• Newton’s Laws are vector equations
• Sometimes easier to relate speed of a
particle to how far it moves under a
force – a single equation can be used –
need to introduce concept of work
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What is work?
• Assume constant force in 1D
• Consider:
vF2 = vI2 + 2as
• Multiply by m/2 
(1/2)mvF2 - (1/2)mvI2 = mas
• But: F = ma
 (1/2)mvF2 - (1/2)mvI2 = Fs
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Work-Kinetic Energy theorem (1)
(1/2)mvF2 - (1/2)mvI2 = Fs
Points:
• W = Fs  defines work done on
particle
= force times displacement
• K = (1/2)mv2  defines kinetic energy
=1/2 mass times square of v
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Work-Kinetic Energy demo
• Cart, force probe, and motion detector
• Plot v2 vs. x – gradient 2F/m
• constant F (measure) -- pulling with string
• Weigh cart and masses in advance
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Conclusions from experiment
Although the motion of the two carts looks very different
(i.e., different amounts of time, accelerations, and final
speeds), there is a quantity that is the same for both at
the end of the motion. It is (1/2) mv2 and is called the
(final) kinetic energy of the carts.
Moreover, this quantity happens to have the same value
as F Ds, which is given the name work.
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Improved definition of work
• For forces, write F  FAB
• Thus W = Fs  WAB = FAB DsA is
work done on A by B as A undergoes
displacement DsA
• Work-kinetic energy theorem:
Wnet,A = SBWAB = DK
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Non-constant force …
Work done in small interval Dx
DW = F Dx
F
Total W done from A to B
S F Dx =
Area under curve!
F(x)
A Dx
Physics 215 – Fall 2014
B
x
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The Work - Kinetic Energy Theorem
Wnet = DK = Kf - Ki
The net work done on an object is equal to
the change in kinetic energy of the object.
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Suppose a tennis ball and a bowling ball are
rolling toward you. The tennis ball is moving
much faster, but both have the same momentum
(mv), and you exert the same force to stop each.
Which of the following statements is correct?
1. It takes equal distances to stop each ball.
2. It takes equal time intervals to stop each ball.
3. Both of the above.
4. Neither of the above.
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Suppose a tennis ball and a bowling ball are
rolling toward you. Both have the same
momentum (mv), and you exert the same force to
stop each.
It takes equal time intervals to stop each ball.
The distance taken for the bowling ball to stop is
1. less than.
2. equal to
3. greater than
the distance taken for the tennis ball to stop.
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Two carts of different mass are accelerated from rest
on a low-friction track by the same force for the same
time interval.
Cart B has greater mass than cart A (mB > mA). The final
speed of cart A is greater than that of cart B (vA > vB).
After the force has stopped acting on the carts, the
kinetic energy of cart B is
1.
2.
3.
4.
less than the kinetic energy of cart A (KB < KA).
equal to the kinetic energy of cart A (KB = KA).
greater than the kinetic energy of cart A (KB > KA).
“Can’t tell.”
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Revised definitions for
Work and Kinetic Energy
Work done on object 1 by object 2:
W(on 1 by 2) = F1,2 ·Dsof 1
Kinetic energy of an object:
K = mv [or:
1
2
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2
1
2
m(v·v)]
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Scalar (or “dot”) product of vectors
The scalar product is a way to combine two
vectors to obtain a number (or scalar). It is
indicated by a dot (•) between the two vectors.
(Note: component of A in direction n is just A•n)
A · B = AB cosq
= Ax Bx + Ay By + Az Bz
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Is the scalar (“dot”) product
of the two vectors
1.
2.
3.
4.
positive
negative
equal to zero
“Can’t tell.”
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Is the scalar (“dot”) product
of the two vectors
1.
2.
3.
4.
positive
negative
equal to zero
“Can’t tell.”
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Two identical blocks slide down two frictionless ramps.
Both blocks start from the same height, but block A is on
a steeper incline than block B.
The speed of block A at the bottom of its ramp is
1.
2.
3.
4.
less than the speed of block B.
equal to the speed of block B.
greater than the speed of block B.
“Can’t tell.”
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Solution
• Which forces do work on block?
• Which, if any, are constant?
• What is F•Ds for motion?
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Work done by gravity
Work W = -mg j•Ds
N
Ds
Therefore,
W = -mgDh
mg
j
N does no work!
i
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Reading assignment
• More W-KE Theorem
• Conservative and non-conservative
forces
• Power
• Finish chapter 11 in textbook
Physics 215 – Fall 2014
Lecture 08-2
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