Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m...

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Solution: Solution: Physics 1710 Physics 1710 Chapter 7&8—Power & Energy Chapter 7&8—Power & Energy K = ½ mv K = ½ mv 2 2 = ½ (2.0 kg) (5.0 m/s) = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m = 25. kg m 2 /s /s 2 = 25. J = 25. J

Transcript of Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m...

Page 1: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Solution:Solution:

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

K = ½ mv K = ½ mv 2 2

= ½ (2.0 kg) (5.0 m/s) = ½ (2.0 kg) (5.0 m/s) 22

= 25. kg m= 25. kg m22/s/s22

= 25. J= 25. J

Page 2: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

What is the minimum height What is the minimum height from which a small rolling ball from which a small rolling ball must be started from rest so that must be started from rest so that it will complete a loop-the-loop?it will complete a loop-the-loop?

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

RR

hh

Page 3: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

No Talking!No Talking!Think!Think!

Confer!Confer!

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What is the minimum height What is the minimum height from which a small rolling ball from which a small rolling ball must be started from rest so must be started from rest so that it will complete a loop-the-that it will complete a loop-the-loop?loop?

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 4: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

What is the minimum height What is the minimum height from which a small rolling ball from which a small rolling ball must be started from rest so that must be started from rest so that it will complete a loop-the-loop?it will complete a loop-the-loop?

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

RR

hh

vv22/R = g/R = g

K = U K = U

½ mv ½ mv 22 = mg(h-R) = mg(h-R)

v v 22 = 2 g (h-R) = 2 g (h-R)

g R = 2g (h-R)g R = 2g (h-R)

h = 3Rh = 3R

vv

Page 5: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

What is the minimum height What is the minimum height from which a small rolling from which a small rolling ball must be started from ball must be started from rest so that it will complete a rest so that it will complete a loop-the-loop?loop-the-loop?

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

K = ½ mv K = ½ mv 2 2

= ½ (2.0 kg) (5.0 m/s) = ½ (2.0 kg) (5.0 m/s) 22

= 25. kg m= 25. kg m22/s/s22

= 25. J= 25. J

Page 6: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

11′′ Lecture Lecture • Power is the time rate of change in Power is the time rate of change in energy. energy.

[Power]= [Watt] = [Joule]/[second][Power]= [Watt] = [Joule]/[second]

• Potential Energy Potential Energy UU is the energy stored is the energy stored in a in a system and may later produce work.system and may later produce work.

• The Potential Energy is equal to the The Potential Energy is equal to the negative of negative of the work done on the the work done on the system to put it in its system to put it in its present state.present state.

• The sum of all energy, potential and The sum of all energy, potential and kinetic, is kinetic, is conserved in an isolated system.conserved in an isolated system.

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 7: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Power:Power:

P P = = dE/dtdE/dt

• Power is the time rate of change in the Power is the time rate of change in the energy of a system, the rate of work down energy of a system, the rate of work down on or by the system.on or by the system.

•Unit of power = Joule/second = WattUnit of power = Joule/second = Watt

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 8: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Unit of Work and Energy:Unit of Work and Energy:

[[F F ‧ ‧ d d ] = N] = N‧m = Joule = ‧m = Joule = JJ

JouleJoule

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 9: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Power: Power: Watt = Joule/secondWatt = Joule/second

James WattJames Watt

Watt’s Steam Engine Watt’s Steam Engine 17741774

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 10: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Power:Power:

P P = = dE/dtdE/dt

P P = = ∆∆E/E/∆∆tt

∆∆E E = = PP ∆∆tt

∆∆E E ==(100 W)(3600 s) (100 W)(3600 s)

∆∆E E ==360 000 J = 360 kJ360 000 J = 360 kJ

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 11: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Potential Energy:Potential Energy:

W = W = ∫ ∫ F•F•d d rr

U = -WU = -W

• Potential Energy is the negative of the work Potential Energy is the negative of the work required to put the system in the current state.required to put the system in the current state.

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 12: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Potential Energy:Potential Energy:

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

FF

-F-F

hh

U = U = - (- F h)- (- F h)

= = m g hm g h

Page 13: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Example: Elevated Mass Example: Elevated Mass

F = -mg F = -mg • Potential Energy:Potential Energy:

UUgg = - = -∫∫00

hhFdy = Fdy = --∫∫00

hh(- mg) dy(- mg) dy

UUgg = mg∫ = mg∫00

h h dy = mgh dy = mgh

• Thus, the potential energy stored in an elevated Thus, the potential energy stored in an elevated mass is proportional to the mass is proportional to the height hheight h and the and the weightweight of the mass.of the mass.

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 14: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

No Talking!No Talking!Think!Think!

Confer!Confer!

Peer Instruction Peer Instruction TimeTime

Where does the energy come from to Where does the energy come from to produce electrical power in a produce electrical power in a hydroelectric dam?hydroelectric dam?

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 15: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Potential Energy:Potential Energy:

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

FF

-F-F

hh

U = U = m g hm g h

P P = = dU/dt dU/dt

= mg dh/dt= mg dh/dtmg =(100. mg =(100. kg)(9.8N/kg)kg)(9.8N/kg)

= 98.0 N= 98.0 N

dh/dt dh/dt = 10 m/10 = 10 m/10 s s

= 1 m/s= 1 m/s

P P = 98. W= 98. W

Page 16: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Relationship Between Relationship Between F F and and U:U:

U = -U = -∫ ∫ F•F•dd r r SoSo

U = -∫ [ FU = -∫ [ Fx x dx + Fdx + Fyy dy + F dy + Fzz dz] dz]ThenThen

FFx x =-dU/dx ; F=-dU/dx ; Fyy =-dU/dy; F =-dU/dy; Fzz =-dU/dz =-dU/dz

FF = - = -∇∇UU

F= -gradient of F= -gradient of UU

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 17: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Example: Mass on a SpringExample: Mass on a Spring

Potential Energy:Potential Energy:

U = ½ k x U = ½ k x 22

F =dU/dxF =dU/dxF= -½ k dxF= -½ k dx22/dx/dx

F= -k xF= -k x

• Thus, the force is equal to the negative of the gradient of Thus, the force is equal to the negative of the gradient of the potential energy.the potential energy.

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 18: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

The FThe Force is equal to the negative orce is equal to the negative gradient of the potential energy:gradient of the potential energy:

FF = - = -∇∇UUFFx x = -∂U/∂x = -∂U/∂x

FFyy = -∂U/∂y = -∂U/∂y

FFzz = -∂U/∂z = -∂U/∂z

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 19: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Physics 1710Physics 1710 Chapter 8 Potential Energy and Chapter 8 Potential Energy and ConservationConservation

Example:Example: Ball on a slopeBall on a slope

• h = ax + byh = ax + by• U = mghU = mgh• FFxx = - = -∂U/∂x = ∂U/∂x = --∂(mgh)/∂x = -mg∂h/∂x ∂(mgh)/∂x = -mg∂h/∂x Similarly:Similarly:

FFyy = = --∂U/∂y = -mg b∂U/∂y = -mg b

• Thus, Thus, F = -F = -mgmg( ( aa i + i + b b j j ))

Page 20: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Conservation of EnergConservation of Energy:y:

• The sum of all energy in a system is The sum of all energy in a system is conserved, i.e. remains the same.conserved, i.e. remains the same.

E = U + KE = U + K

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 21: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Example: PendulumExample: Pendulum

U = mg hU = mg hh = L(1- cos h = L(1- cos ))

U = mg L(1- cos U = mg L(1- cos ))

K = ½ m v K = ½ m v 22

=½ m (Ld =½ m (Ld /dt) /dt) 22

E = mg L(1- cos E = mg L(1- cos ) + ½ m (Ld ) + ½ m (Ld //dt) dt) 2 2

= constant= constant

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 22: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Thought (Thought (Gedanken) Gedanken) Experiment:Experiment:

• Why does a pendulum stop Why does a pendulum stop moving?moving?

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 23: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Dissipative (non-conservative) Forces:Dissipative (non-conservative) Forces:

W = W = ∫ ∫ F•F•d d rr

==∫ (C v∫ (C vxx 2 2 )dx)dx

==∫ (C v∫ (C vxx 2 2 )(dx /dt) dt)(dx /dt) dt

=∫ (C v=∫ (C vxx 3 3 )dt)dt

E = U + K -WE = U + K -W

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy

Page 24: Solution: Physics 1710 Chapter 7&8—Power & Energy K = ½ mv 2 = ½ (2.0 kg) (5.0 m/s) 2 = 25. kg m 2 /s 2 = 25. J.

Summary:Summary:

•The Potential Energy is equal to the The Potential Energy is equal to the negative of the work done on the system to negative of the work done on the system to put it in its present state.put it in its present state.

U = -U = -∫ F•d r∫ F•d r• The sum of all energy, potential and The sum of all energy, potential and kinetic, of a system is conserved, in the kinetic, of a system is conserved, in the absence of dissipation.absence of dissipation.

E = U + K – WE = U + K – W• F = - F = - ∇U∇U

•P P = = dE/dtdE/dt

Physics 1710Physics 1710 Chapter 7&8—Power & Chapter 7&8—Power & EnergyEnergy