As 21d Workenergy&Power

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2.1d Mechanics Work, energy and power Breithaupt pages 148 to 159  April 14 th , 2012

Transcript of As 21d Workenergy&Power

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2.1d Mechanics 

Work, energy and powerBreithaupt pages 148 to 159

 April 14th, 2012

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 AQA AS Specification

Lessons  Topics 

1 & 2 Work, energy and power 

W = Fs cos θ  

 P = ΔW / Δt

 P = Fv 

3 & 4 Conservation of energy Principle of conservation of energy, applied to examples involving gravitational

 potential energy, kinetic energy and work done against resistive forces.

 ΔE  p = mgΔh 

 E k  = ½ mv2 

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Work (W )

Work is done when a force moves its point ofapplication.

work  = force  x distance moved in thedirection of the force

W = F s

unit: joule (J)work is a scalar  quantity

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If the direction of the force and the distance

moved are not in the same direction:

W = F s cos θ  

The point of application of force, F  moves

distance s cos θ  when the object moves

through the distance s .

F

s

θ  object

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Question 1

Calculate the work done when a force of5 kN moves through a distance of 30 cm

work = force x d istance= 5 kN x 30 cm

= 5000 N x 0.30 m

work = 1500 J

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Question 2

Calculate the work done by a child of weight 300N who

climbs up a set of stairs consisting of 12 steps each of

height 20cm.

wo rk = force x distance

the child must exert an upward force equal to its weight

the distance moved upwards equals (12 x 20cm) = 2.4m

work = 300 N x 2.4 m

work = 720 J

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Question 3

Calculate the work done by

the wind on the yacht in the

situation shown below:

W = F s cosθ 

 = 800 N x 50 m x cos

30° 

= 40 000 x cos 30° 

= 40 000 x 0.8660work = 34 600 J

wind force = 800 N

distance moved

by yacht = 50 m

30° 

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Complete:

Force Distance Angle between

F  and sWork

400 N 5 km 0°  2 MJ

200 μN 300 m 0°  60 mJ

50 N 6 m 60°  150 J

400 N 3 m 90°  0 J

 Answers

400 N

300 m

60° 

0 J *

* Note:  No work is done when the force and

distance are perpendicular to each other.

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Force-distance graphs

The area under thecurve is equal to

the work done.

F

s

force

distance

area = work done

F

s

force

distance

area = work

= ½ F s

area = work

found by

countingsquares on

the graph 

F

s

force

distance

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Question

Calculate the work done by

the brakes of a car if theforce exerted by the brakesvaries over the car’s brakingdistance of 100 m as shownin the graph below. 

Work = area under graph  

= area A + area B

= (½ x 1k x 50)

+ (1k x 100)

= (25k) + (100k)

work = 125 kJ

force / kN

distance / m

50 100

area B

area A

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Energy (E )

Energy is needed to move objects, to changetheir shape or to warm them up.

Work is a measurement of the energy required to

do a particular task.

work done = energy change

unit: joule (J)

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Conservation of Energy

The principle of the conservation ofenergy states that energy cannot be

created or destroyed.

Energy can change from one form to

another.

 All forms of energy are scalar quantities 

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Some examples of forms of energyKinetic energy (KE)

Energy due to a body’s motion. Potential energy (PE)

Energy due to a body’s position

Thermal energy

Energy due to a body’stemperature.

Chemical energy

Energy associated with chemical

reactions.

Nuclear energy

Energy associated with nuclearreactions.

Electrical energy

Energy associated with electric

charges.

Elastic energy

Energy stored in an object when it

is stretched or compressed.

All of the above forms of energy (and others) can

ultimately be considered to be variations of kinetic or

potential energy.

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Kinetic Energy (E K )

Kinetic energy is the energy an object hasbecause of its motion and mass.

kin etic energy = ½ x mass x (speed) 2  

E K  = ½ m v 2

Note: v  = speed NOT velocity.

The direction of motion has no relevance to kineticenergy.

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Question 1

Calculate the kinetic energy of a car of mass800 kg moving at 6 ms-1

E K  = ½ m v 2

= ½ x 800kg x (6ms-1)2

= ½ x 800 x 36

= 400 x 36 

kinetic energy = 14 400 J

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Question 2

Calculate the speed of a car of mass 1200kg if itskinetic energy is 15 000J

E K  = ½ m v 2

15 000J = ½ x 1200kg x v 2  

15 000 = 600 x v 

 15 000 ÷ 600 = v 2

25 = v 2  

v  = 25

speed = 5.0 ms-1

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Question 3Calculate the braking

distance a car of mass900 kg travelling at aninitial speed of 20 ms-1 ifits brakes exert a constantforce of 3 kN.

k.e. of car = ½ m v 2

= ½ x 900kg x (20ms-1)2

= ½ x 900 x 400 

= 450 x 400 

k.e. = 180 000 J

The work done by the

brakes will be equal to thiskinetic energy.

W = F s

180 000 J = 3 kN x s

180 000 = 3000 x ss  = 180 000 / 3000 

braking distance = 60 m

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Complete:

Mass Speed K inetic energy

400 g 4.0 ms-1 3.2 J

3000 kg 10 kms-1 60 mJ

8 kg 300 cms-1 36 J

50 mg 12 ms-1 3.6 mJ

 Answers

8 kg

12 ms-1

1.5 x 1011 J

3.2 J

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Gravitational Potential Energy (gpe )

Gravitational potential energy is theenergy an object has because of itsposition in a gravitational field.

change in g .p .e.

= mass x gravi tat ional f ield strength

x change in height

 ΔE P  = m g  Δh

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Question

Calculate the change in g.p.e. when a massof 200 g is lifted upwards by 30 cm.

(g = 9.8 Nkg -1 ) 

 ΔE P  = m g  Δh= 200 g x 9.8 Nkg-1 x 30 cm 

= 0.200 kg x 9.8 Nkg-1 x 0.30 m 

change in g.p.e. = 0.59 J

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Complete:

mass g Δh  

 ΔE P

3 kg 10 Nkg-1 400 cm 120 J

200 g 1.6 Nkg-1 30 m 9.6 J

7 kg 10 Nkg-1 4000 m 280 kJ

2000 g 24 Nkg-1 3000 mm 144 J

 Answers

3 kg

1.6 Nkg-1

4000 m

144 J

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Falling objects

If there is no significant

air resistance then the

initial GPE of an object

is transferred into

kinetic energy.

 ΔE K  =  ΔE P  

½ m v 2 = m g  Δh  

 Δh

m

½ Δh

v 1

v 2

gpe = mg  Δh

ke = ½ mv 2 2 

 

ke = 0  

gpe = 0

gpe = ke

gpe = ½ mg  Δh

ke = ½ mv 1 2  

ke = mg  Δh

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Question A child of mass 40 kgclimbs up a wall of height2.0 m and then steps off. Assuming no significantair resistance calculate themaximum:

(a) gpe of the child

(b) speed of the child

g  = 9.8 Nkg -1

(a) max gpe occurs whenthe child is on the wall

gpe = mg  Δh

= 40 x 9.8 x 2.0

max gpe = 784 J

(b) max speed occurs whenthe child reaches the ground

½ m v 2 = m g  Δh½ m v 2 = 784 Jv 2 = (2 x 784) / 40

v 2 = 39.2v  = 39.2

max speed = 6.3 ms-1

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Power (P )

Power is the rate of transfer of energy.

power = energy transfer

t imeP =  ΔE

 Δt

unit: watt (W)

power is a scalar quantity

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Power is also the rate of doing work.

power = work done

t ime

P =  ΔW Δt  

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Question 1

Calculate the power of an

electric motor that lifts amass of 50 kg upwards by

3.0 m in 20 seconds.

g  = 9.8 Nkg -1 

 ΔE P  = m g  Δh

= 50 kg x 9.8 Nkg-1 x 3 m

= 1470 J

P = ΔE /

 Δt  

= 1470 J / 20 s

power = 74 W

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Question 2Calculate the power of a car engine that exerts a force of

40 kN over a distance of 20 m for 10 seconds.

W = F s

= 40 kN x 20 m

= 40 000 x 20 m= 800 000 J

P =  ΔW /  Δt  

= 800 000 J / 10 spower = 80 000 W

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Complete:

energy

transfer

wo rk done t ime   power  

600 J 600 J  2 mins 5 W

440 J 440 J 

20 s 22 W

28 800 J 28 800 J 2 hours 4 W

2.5 mJ 2.5 mJ 50 μs 50 W

 Answers

600 J 5 W 

440 J 20 s28 800 J 28 800 J 

2.5 mJ 50 W

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Power and velocity

power = work done / t ime

but: work = force x disp lacement

therefore: power = force x disp lacement

t ime

but: disp lacement / t ime = veloci ty

therefore:

power = force x veloc i tyP = F v

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Question

Calculate the power of a car

that maintains a constantspeed of 30 ms-1 against air

resistance forces of 2 kN

 As the car is travelling at a

constant speed the car’sengine must be exerting a

force equal to the opposing

air resistance forces.

P = F v

= 2 kN x 30 ms-1 

= 2 000 N x 30 ms-1 

power = 60 kW

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Energy efficiency

Energy efficiency is a measure of howusefully energy is used by a device.

efficiency =

useful energy transferred by the device

total energy supplied to the device

 As the useful energy can never be greater

than the energy supplied the maximumefficiency possible is 1.0

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 Also:

efficiency =useful work output

energy supplied

useful power outputefficiency =

power input

In all cases:

percentage efficiency = efficiency x 100

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Complete

Input

energy (J)

Useful

energy (J)

Wasted

energy (J)

Efficiency Percentage

efficiency

100 40

250 50

50 0.20

80 30%

60 60

60

200

10 40

24 56

120

0.80

0.50

0.30

20%

0.40

80%

50%

40%

Answers

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Internet Links• Reaction time stopping a car  - also plots velocity/time graph - NTNU

•Car Accident & Reaction Time - NTNU

• Work  (GCSE) - Powerpoint presentation by KT

• Kinetic Energy (GCSE) - Powerpoint presentation by KT

• Gravitational Potential Energy (GCSE) - Powerpoint presentation by KT

• Energy Skate Park  - Colorado - Learn about conservation of energy with a

skater dude! Build tracks, ramps and jumps for the skater and view the

kinetic energy, potential energy and friction as he moves. You can also take

the skater to different planets or even space!

• Rollercoaster Demo - Funderstanding

• Energy conservation with falling particles - NTNU

• Ball rolling up a slope- NTNU

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Core Notes from Breithaupt pages 148 to 159

1. What is the principle ofconservation of energy?

2. Define work and give its unit.Explain how work iscalculated when force anddistance are not in the samedirection.

3. With the aid of a diagramexplain how work can befound from a graph.

4. Explain what is meant by,and give equations for (a)kinetic energy & (b)gravitational potential energy.

5. In terms of energy explainwhat happens as a body fallsunder gravity.

6. In terms of energy and workdefine power.

7. Show that the power of anengine is given by: P = Fv .

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Notes from Breithaupt pages 148 to 150

Work and energy

1. What is the principle of conservation of

energy?

2. Define work and give its unit. Explain how work

is calculated when force and distance are notin the same direction.

3. With the aid of a diagram explain how work can

be found from a graph.

4. Try the summary questions on page 150

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Notes from Breithaupt pages 151 & 152

Kinetic and potential energy

1. Explain what is meant by, and give equationsfor (a) kinetic energy & (b) gravitationalpotential energy.

2. In terms of energy explain what happens as abody falls under gravity.

3. Repeat the worked example on page 152 this

time where the track drops vertically 70 m andthe train has a mass of 3000 kg.

4. Try the summary questions on page 152  

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Notes from Breithaupt pages 153 & 154

Power

1. In terms of energy and work define power.

2. Show that the power of an engine is given by:

P = Fv .

3. Repeat the worked example on page 154 this

time where the engine exerts a force of 50 kN

with a constant velocity of 100 ms-1.

4. Try the summary questions on page 154 

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Notes from Breithaupt pages 155 & 156

Energy and efficiency

1. Try the summary questions on page 156  

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Notes from Breithaupt pages 157 to 159

Renewable energy

1. Try the summary questions on page 159