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Newton’s 3rd Law
Momentum
Prentice Hall
Chapters 7 & 8
1
Ch 7 & 8 assignments
– Complete Packets for the chapters. On the
problem sections, you must follow the
problem solving method: write givens with
symbols and units, working equation, work
with units and answer with units. Use more
paper if there is not enough room.
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Newton’s 3rd Law
• For every action there is an opposite and
equal reaction.
• This is applied to momentum and
collisions.
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Chapter 8 Momentum and Its
Conservation
Momentum, r = mass of an object times
its velocity
r = mv
units: kgm/s
Impulse is an object’s difference in final
momentum and initial momentum
F D t = m D v = r f– r i
units: ns, kgm/s
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Conservation of Momentum
This is used in collisions
where Newton’s third law
is related to conservation
of momentum.
See figure 8.11 on page 132
ri = r f
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6
Recoil occurs when objects are at
rest initially.
Example problem on page 135
General formula
r i = r f but r I (small fish) = 0
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Collision Types
• Elastic – when objects collide and
rebound from each other without a loss in
energy (no permanent change in shape
and no heat change). Fig 8.11 (pg. 132)
• Inelastic – when objects collide and
become distorted. Vehicle crashes.
• Perfectly inelastic – when objects collide
and stick together after impact. Freight
cars coupling on train track.
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Asgn:
• 142/55,57,58
due tomorrow.
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10
Now we begin:
Work
Power
Energy
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Objectives
*Differentiate among the various forms of
energy and how they can be transformed
from one into another.
*Explore the Law of Conservation of Energy
by differentiating among open, closed, and
isolated systems and explain that the total
energy in an isolated system is a conserved
quantity.
*Compare and contrast work and power
both qualitatively and quantitatively.
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Work is the transfer of energy by
mechanical means.
• In this section you will calculate work
and power used.
• Terms/Units:
Work/joule,j energy/joule,j
kinetic energy/joule,j
work-energy theorem
Power/joule/sec or watt
746 w = 1 hp
13
Work = Force x distance
• W = Fd
units: nm or joule, j
• Work is equal to a constant force exerted
on an object in the direction of motion,
times the object’s displacement
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Kinetic Energy – the energy of
motion
• KE = ½ m v 2
• units: kg (m/s) 2 = (kgm/s2) m
•
= nm = joule
• Work Energy Theorem: W = D KE
• Work = the change in kinetic energy of a
system.
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Work at an angle
• W = Fd cos q
• Work = the product of force and
displacement, times the cosine of the
angle between the force and the direction
of the displacement.
• How much work is done pulling with a 15
N force applied at 20.o over a distance of
12 m?
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How much work is done pulling with a 15 N
force applied at 20.o over a distance of 12
m?
20o
• W = Fdcos q
• W = 15 N (12 m) (cos 20)
• W = 169  170 J
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Power
Power = Work / time
j/s = watt or w
Power is “The rate of doing work.”
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Sample Problem:
• Together, two students exert a force of
825 n in pushing a car a distance of 35 m.
• A. How much work do they do on the car?
• W = Fd = 825n (35m) =
• B. If the force was doubled, how much
work would they do pushing the car the
same distance?
• W = 2Fd = 2(825n)(35m) =
19
A rock climber wears a 7.5 kg backpack while
scaling a cliff. After 30.0 min, the climber is 8.2
m above the starting point.
a.How much work does the climber do on the
backpack?
W = Fd = mgd = 7.5kg(9.8m/s2)(8.2m)
b. How much power did the climber expend?
P = W = 6.0x102 j x 1min = 0.33 = 3.3x10-1 w
t
30.0min
60sec
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Machines
21
Actual Mechanical Advantage –
comparing torques or forces
MA = resistance (or load) force
effort force
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• Calculated Actual Mechanical Advantage:
• Fr / Fe = 200nm / 300 nm = .67
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Ideal Mechanical Advantage, comparing
distances:
IMA = d effort / d resistance
IMA = 2.4 m / 1.2 m = 2
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Efficiency
Efficiency, e = work out = AMA x 100
work in
IMA
Sample problem on next slide.
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Two people use a wheel and axle to raise a mass
of 750. kg. The radius of the wheel is 0.50 m and
the radius of the axle is 0.040 m. If the efficiency of
the machine is 62% and each person exerts and
equal force, how much force must each apply?
Efficiency, e = work out = AMA x 100
work in
IMA
e = 0.62 IMA = R =0.50m =
r 0.040m
AMA = f effort
 f effort = f resist AMA  f effort = ?
f resistance
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Efficiency, e = work out = AMA x 100
work in
IMA
e = 0.62 IMA = R =0.50m =
r 0.040m
AMA = f effort  f effort = f resist AMA  f effort = ?
fresistance
f effort = f resis r = 750. kg x 10.nkg-1 x 0.040m
e xR
0.62 x 0.50 m
f effort = 968 n
Per person: 968 = 484 n
2
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Chapter 11 Energy & Its
Conservation
• Kinetic Energy: KE = ½ m v2
• KE for a Spring: KE = ½ k (Dd)2
k=F/d
• Potential Energy: PE = mgh
• Rest Energy: E = mc2
• Mechanical Energy, E = KE + PE
• Conservation of Energy (collisions, etc.)
• KEi + PEi = KEf + PEf
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The total energy of a closed, isolated
system is constant. The energy can
change form.
20.0J
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Examples: Collisions
• Elastic Collisions: ones in which initial KE
= final KE
• Inelastic Collisions: ones in which energy
is changed into another form (i.e., KE into
heat or sound)
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A 98.0 –N sack of grain is hoisted to a
storage room 50.0 m above the ground floor
of a grain elevator.
a. How much work was done?
W = Fd
W = 98.0 – N ( 50.0m)
W = 4900 = 4.90 x 103 j
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A 98.0 –N sack of grain is hoisted to a
storage room 50.0 m above the ground floor
of a grain elevator.
b. What is the increase in potential energy of
the sack of grain at this height?
PE = mgh = Fd
PE = 98.0 N (50.0 m)
PE = 4900 j = 4.90 x 103 j
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A 98.0 –N sack of grain is hoisted to a
storage room 50.0 m above the ground floor
of a grain elevator.
c. The rope being used to lift the sack of
grain breaks just as the sack rached the
storage room. What kinetic energy does
the sack have just before it strikes the
ground floor?
Ans.:
KEbottom = PEtop
KEbottom = 4.90 x 103 J
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11/73
73a. W = Fd = 98.0 N (50.0 m)
W = 4.90 x 103 J
73b. PE = mgh = Fd
DPE = W = 4.90 x 103 J
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