The Work-Energy Theorem Derivation: v f 2 =v i 2 +2ad Kinematics Equation mv f 2 =mv i 2...

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The Work-Energy The Work-Energy Theorem Theorem Derivation: Derivation: v v f 2 =v =v i 2 +2ad +2ad Kinematics Kinematics Equation Equation mv mv f 2 =mv =mv i 2 +2mad +2mad Dynamics equation Dynamics equation (1/2)[mv (1/2)[mv f 2 =mv =mv i 2+2mad] 2+2mad] mad= ½ mv mad= ½ mv f 2 - ½ mv - ½ mv i 2 KE=kinetic energy = ½ mv 2 Kinetic Energy: The energy associated with the motion of a mass The amount of capable work associated with the motion of a mass.

Transcript of The Work-Energy Theorem Derivation: v f 2 =v i 2 +2ad Kinematics Equation mv f 2 =mv i 2...

Page 1: The Work-Energy Theorem Derivation: v f 2 =v i 2 +2ad Kinematics Equation mv f 2 =mv i 2 +2madDynamics equation (1/2)[mv f 2 =mv i 2+2mad] mad= mv f.

The Work-Energy The Work-Energy TheoremTheoremDerivation:Derivation:

vvff22=v=vii

22+2ad +2ad Kinematics Kinematics EquationEquationmvmvff

22=mv=mvii22+2mad+2mad Dynamics Dynamics

equationequation(1/2)[mv(1/2)[mvff

22=mv=mvii2+2mad]2+2mad]mad= ½ mvmad= ½ mvff

22- ½ mv- ½ mvii22

KE=kinetic energy = ½ mv2

Kinetic Energy: The energy associated with the motion of a mass The amount of capable work associated with the motion of a mass.Measured in Joules (J)KE is a scalar quantity.

Page 2: The Work-Energy Theorem Derivation: v f 2 =v i 2 +2ad Kinematics Equation mv f 2 =mv i 2 +2madDynamics equation (1/2)[mv f 2 =mv i 2+2mad] mad= mv f.

Kinetic Energy Kinetic Energy ConsiderationsConsiderations

A small mass moving at a high speed can have the same kinetic energy (do the same work) as a large mass that travels slowly.

You have two equivalent masses, mass A and B. Suppose you double the mass of A and double the speed of mass B. Which mass now has more KE?

Why? A car traveling at a certain speed stops after travel a

distance, d, after the brakes are applied. Suppose that the speed of the car is double, by what factor does the distance to stop change provided that the braking for is the same as the first situation.

Page 3: The Work-Energy Theorem Derivation: v f 2 =v i 2 +2ad Kinematics Equation mv f 2 =mv i 2 +2madDynamics equation (1/2)[mv f 2 =mv i 2+2mad] mad= mv f.

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Page 4: The Work-Energy Theorem Derivation: v f 2 =v i 2 +2ad Kinematics Equation mv f 2 =mv i 2 +2madDynamics equation (1/2)[mv f 2 =mv i 2+2mad] mad= mv f.

Things for ThoughtThings for Thought A 2 g bumblebee moving at 2 m/s has a

kinetic energy of 4x10-3 J which means it is capable of doing 4x10-3 J of work

A comet with a mass of 7.9x1011 kg traveling at 25,000 m/s has a kinetic energy or capable of doing 2.5x1020J of work. The most powerful atomic bomb has an energy yield of 5x1015 J, thus this comet can do 5000 times more work.

Page 5: The Work-Energy Theorem Derivation: v f 2 =v i 2 +2ad Kinematics Equation mv f 2 =mv i 2 +2madDynamics equation (1/2)[mv f 2 =mv i 2+2mad] mad= mv f.

Various Forms of the Work-Energy Various Forms of the Work-Energy TheoremTheorem

mad= ½ mvmad= ½ mvff22- ½ mv- ½ mvii

22

(Fnet)d=KE(Fnet)d=KEff-KE-KEii KEKEff=final kinetic energy, KE=final kinetic energy, KEii=initial kinetic =initial kinetic energyenergy

Wnet=KEWnet=KEff – KE – KEiiWnet=Wnet=ΔΔKEKE The net work accomplished on mass causes a change in The net work accomplished on mass causes a change in

its kinetic energy.its kinetic energy. The work-energy theorem is a cause and effect The work-energy theorem is a cause and effect

relationship. The net work on a mass causes a change relationship. The net work on a mass causes a change in its kinetic energy.in its kinetic energy.

Positive Net Work – increases the speed of a mass Positive Net Work – increases the speed of a mass causing a positive change in kinetic energy.causing a positive change in kinetic energy.

Net Negative Work – decreases the speed of a mass Net Negative Work – decreases the speed of a mass causing a negative change in kinetic energy.causing a negative change in kinetic energy.


Momentum and Its Conservation · a. F!t! p f $ p i! mv f $ mv i v f!!! 2.7 m/s in the same direction as the original velocity b. v f!!! 1.3 m/s in the same direction as the original

Momentum and Its Conservation · a. F!t! p f $ p i! mv f $ mv i v f!!! 2.7 m/s in the same direction as the original velocity b. v f!!! 1.3 m/s in the same direction as the original

Infiniium MXR-Series Data Sheet · 100 mV/div 681 µV 821 µV 1.06 mV 1.17 mV 1.46 mV 2.03 mV 1 V/div 5.07 mV 6.33 mV 8.4 mV 9.31 mV 11.91 mV 16.26 mV ENOB on 50 Ω inputs, 50 mV/div,

Infiniium MXR-Series Data Sheet · 100 mV/div 681 µV 821 µV 1.06 mV 1.17 mV 1.46 mV 2.03 mV 1 V/div 5.07 mV 6.33 mV 8.4 mV 9.31 mV 11.91 mV 16.26 mV ENOB on 50 Ω inputs, 50 mV/div,

MV 2-22-13

MV 2-22-13

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Meets 100% now it’s time to take your students there! of theergopedia.com/ergoweb/img/home/home/EP_2014_ok.pdfW W x k m x F = mgh +1 2 mv 2+1 2 kx gh +1 2 2 mv 2+ 1 _ 2 kx = 1 _

Meets 100% now it’s time to take your students there! of theergopedia.com/ergoweb/img/home/home/EP_2014_ok.pdfW W x k m x F = mgh +1 2 mv 2+1 2 kx gh +1 2 2 mv 2+ 1 _ 2 kx = 1 _

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-140 mV to -80 mV -140 mV to -60 mV -140 mV to -50 mV -140 mV to -100 mV (n=12 ) (n=9) (n=20 ) (n=26 ) (n=12 ) # of channels = 8.

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USV 3200HP 45M (MV Kelisamas 2)

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Mv analysis 2

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6.1. 6.2 6.3 6.4 Inertia, momentum 6.5 (mv) = F t MomentumImpulse (Ft)

6.1. 6.2 6.3 6.4 Inertia, momentum 6.5 (mv) = F t MomentumImpulse (Ft)

7.1. 7.2 7.3 7.4 W = F x d W = F x 0 = 0 7.5 Power = work / time PE KE PE = mgh KE = ½ mv 2.

7.1. 7.2 7.3 7.4 W = F x d W = F x 0 = 0 7.5 Power = work / time PE KE PE = mgh KE = ½ mv 2.

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2-Part1 MV Distribution,2014

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