1 Work, Power, Energy Glencoe Chapters 9,10,11. 2 Ch 9 assignments In class samples: 1,2,4,13,15...
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Transcript of 1 Work, Power, Energy Glencoe Chapters 9,10,11. 2 Ch 9 assignments In class samples: 1,2,4,13,15...
1
Work, Power, Energy
GlencoeChapters 9,10,11
2
Ch 9 assignments
• In class samples:
• 1,2,4,13,15
• Assigned problems
• 7-9,17,20
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Chapter 9 Momentum and Its Conservation
Momentum, = mass of an object times its velocity = mv units: kgm/s
Impulse is an object’s difference in final momentum and initial momentumF t = m v = f – 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 9-6 on page 236
i = f
Ex. prob 2 on p. 237
#2, p237
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Recoil occurs when objects are at rest initially.
Example problem 3 on page 240
General formula
i = f but i = 0
Ci +Di = Cf + Df
0 = Cf + Df
So: Cf = - Df
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8
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Chapter 10 Energy, Work, and Simple Machines
• Assigned problems:
• 29,52,57,60,66,81
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Work is the transfer of energy by mechanical means.
• In this section you will calculate work and power used.
• Terms:
work energy kinetic energy
work-energy theorem
Joule power watt
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Work = Force x distance cos
• 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 = KE
• Work = the change in kinetic energy of a system.
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Work at an angle
• W = Fd cos • 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 20o over a distance of 12 m?
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• W = Fdcos • W = 15 N (12 m) (cos 20)
• W = 169 170 J
20o
15 N
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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: #2 on page 261
• 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) =
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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 does the climber expend in this effort?
P = W = 603 J = 0.34 watt
t 1800 s
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Machines
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Mechanical Advantage
MA = resistance (or load) force effort force
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• Calculated Mechanical Advantage:• Fr / Fe = 200nm / 300 nm = .67
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Ideal Mechanical Advantage,
IMA = d effort / d resistance
IMA = 2.4 m / 1.2 m = 2
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Efficiency
Efficiency, e = work out = MA x 100
work in IMA
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Chapter 11 Energy & Its Conservation
• Kinetic Energy: KE = ½ m v2
• KE for a Spring: KE = ½ k (d)2
k = F / d• Potential Energy: PE = mgh• Rest Energy: E = mc2
• Mechanical Energy, E = KE + PE• Conservation of Energy (collisions, etc.)• KE i + PE i = KE f + PE f
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Ch11 Assignment
• 11/6,55,59,73,74.79
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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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11/73
73a. W = Fd = 98.0 N (50.0 m)
W = 4.90 x 103 J
73b. PE = mgh = Fd
PE = W = 4.90 x 103 J
73c. KEbottom = PEtop
KEbottom = 4.90 x 103 J