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1 3 | 10 http://themalloryfamily.net CHAPTER 4 Work http://themalloryfamily.net 4 | 11 Work Work - the product of the magnitude of the force (F) and the parallel distance (d) through which the object moves work = force x parallel distance W = Fd Mechanically, work involves both force and motion Section 4.1 http://themalloryfamily.net 4 | 12 http://themalloryfamily.net 4 | 13 No work is done because there is no movement (d = 0) Section 4.1 •10 •11 •12 •13

No work is done because there is no movement (d = 0) · 3 4 | 18 Energy • The ability to do work. An object or system that possesses energy has the ability to do work • When work

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Page 1: No work is done because there is no movement (d = 0) · 3 4 | 18 Energy • The ability to do work. An object or system that possesses energy has the ability to do work • When work

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3 | 10http://themalloryfamily.net

CHAPTER 4Work

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Work

• Work - the product of the magnitude of the force (F) and the parallel distance (d) through which the object moves

• work = force x parallel distance• W = Fd• Mechanically, work involves both force and

motion

Section 4.1

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No work is done because there is no movement (d = 0)

Section 4.1

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Work Being Done Applied force (F) through a distance (d)

Section 4.1

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Work - Units

• SI System– W = Fd newton x meter = Nꞏm = joule (J)

• British System– W = Fd pound x foot = foot-pound (ft-lb)

Section 4.1

Nꞏm = joule (J)N = newton = kgꞏm/s2

J = kgꞏm2/s2

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Working against Something

• When work is done, we generally feel the other part of Newton’s third-law force acting against us

• Gravity and friction are common agents working against work

• In the case of gravity, we must apply force to overcome the force of gravity (weight = w = ma)

• Work = W= Fd = wd = mad – (d is the distance lifted)

Section 4.1

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Energy

• The ability to do work. An object or system that possesses energy has the ability to do work

• When work is done by a system, the amount of energy of the system decreases ↓

• When work is done on a system, the system gains energy ↑

• Work is the process by which energy is transferred from one object to another

• Work and Energy have the same units – joules

Section 4.2 http://themalloryfamily.net 4 | 19

Kinetic Energy

• Kinetic Energy - the energy an object possesses because of its motion, the energy of motion

• kinetic energy = ½ x mass x (velocity)2

• Ek = ½mv2

• If an object is already moving• Work = change in kinetic energy• W = k = Ek2 – Ek1 = ½mv2

2 - ½mv12

Section 4.2

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Potential Energy

• Potential Energy - the energy an object has because of its position or location, the energy of position

• Most potential energy is due to gravity• Remember that:

– Work = Force x distance (W = Fd)– Weight is a force (w = ma where a=9.80 m/s2)

• Therefore W = mad– Grav. potential energy = weight x height

• Ep = madSection 4.2 http://themalloryfamily.net 4 | 21

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Gravitational Potential Energy

• The gravitational Potential Energy is equal to the work done and this is equal to the weight times the height

• W = Ep = mad• Example: How much work is done lifting a 1.0

kg book to the height of 1m?• Work = W = (1.0 kg) (9.8 m/s2) (1m) = 9.8 J• This is also the amount of Ep that the book has

at a height of 1m

Section 4.2 http://themalloryfamily.net 4 | 23

Potential Energy

• Depends only on the initial and final positions (difference in height - d) and is independent of path

• If we disregard any frictional loss, it takes the same amount of work (W) to lift the mass (m), no matter the path

Section 4.2

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Work = change in potential energy• Work = W = Ep = mad = mad• In the previous examples the d is actually d• Work is done when there is a change in position• Therefore the reference point for measuring

heights is arbitrary (but must be internally consistent)

Section 4.2 http://themalloryfamily.net 4 | 25

Reference Point No matter which scale – d is the same

Section 4.2

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•24 •25

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Conservation of Energy

• Energy can neither be created nor destroyed.• In changing from one form to another, energy is

always conserved• The total energy of an isolated system remains

constant• (total energy)time1 = (total energy) time2• The total energy does not change with time

Section 4.3 http://themalloryfamily.net 4 | 27

Conservation of Mechanical Energy

• To simplify we will deal with ideal systems – in which energy is only in two forms – kinetic and potential

• Total Energy = Ek + Ep

• Conservation of energy(Ek + Ep)1 = (Ek + Ep)2

(½mv2 + mad)1 = (½mv2 + mad)2

Section 4.3

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Conservation of Energy Finding Kinetic and Potential Energies

• A 0.10 kg stone is dropped from a height of 10.0 m. What will be the kinetic and potential energies of the stone at the heights indicated in the figure (neglect air resistance).

• ET = Ek + Ep will be true at all heights.

• At the moment the stone is released ET= Ep (Ek = 0)

• At the moment the stone hits the ground ET = Ek (Ep = 0)

Section 4.3 http://themalloryfamily.net 4 | 29

Solve for Ep and Ek at heights indicated

• At any height, the potential energy Ep = mad

• d = 10 m: Ep = (0.10 kg)(9.8 m/s2)(10.0 m) = 9.8 J• d = 7 m: Ep = (0.10 kg)(9.8 m/s2)(7.0 m) = 6.9 J• d = 3 m: Ep = (0.10 kg)(9.8 m/s2)(3.0 m) = 2.9 J• d = 0 m: Ep = (0.10 kg)(9.8 m/s2)(0 m) = 0 J• ET = Ek + Ep

• Ek = ET - Ep

Section 4.3

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Solve for Ep and Ek at heights indicated

Height(m)

ET(Joules)

Ep(Joules)

Ek(Joules)

10.0 9.8 9.8 0

7.0 9.8 6.9 2.9

3.0 9.8 2.9 6.9

0 9.8 0 9.8

Section 4.3

ET = Ek + Ephttp://themalloryfamily.net 4 | 31

Magnitude of Velocity

• Potential Energy (Ep) converted = mad• Converted into Kinetic Energy (Ek) = ½mv2

• Since all the Ep is converted into Ek just before hitting the ground, we can use this to compute the speed or magnitude of velocity

• Therefore: ½mv2 = mad• ½v2 = ad (cancel m’s)• v2 = 2ad (solve for v)

Section 4.3

2ad• v =

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Power – SI System

• Power - the time rate of doing work

• SI Units J/s = Watt (1 J/s = 1 W)• For example a 100W light bulbs uses 100

joules/second of electrical power or 100 Watts• Careful not to confuse W (work) with W (watt)

Section 4.4

worktime

Wt

Fdt

• Power = = =

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Force of 1.0 N to raise a mass 1.0 m, the amount of work done is 1.0 J. If this work is done in 1.0 s, then the power is 1.0 W

Section 4.4

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Quantity Unit SymbolEquivalent

Units

Force newton N Kgꞏm/s2

Work joule J Nꞏm

Energy joule J Nꞏm

Power watt W J/s

A Review of SI Units

Section 4.4 http://themalloryfamily.net 4 | 35

Energy

• No matter what type of energy that we speak of – chemical, electrical, mechanical – the main unifying concept is the conservation of energy

• We cannot create or destroy energy, but we can change it from one form to another

Section 4.5

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Forms of Energy

• Thermal (heat) Energy – related to the kinetic and potential energies on a molecular level

• Gravitational potential energy – from an object’s position, stored gravitational energy

• Electrical energy – associated with the motion of electric charges

• Chemical energy – molecular bonds• Radiant energy – sun• Nuclear energy – rearrangement of nuclei

– Fission – breaking apart of larger nuclei– Fusion - smaller nuclei are put together

Section 4.5 http://themalloryfamily.net 4 | 37

Fossil Fuels

• Fossil Fuels – oil, gas, coal - from once living organisms; basically stored solar and chemical energy

• Oil – from marine organisms (U.S. imports more than 50% of our needs)

• Gas – from marine organisms (most is produced domestically)

• Coal – from terrestrial (land) plants (the U.S. has large reserves, but environmental problems in coal mining)

• Methane hydrate – crystalline form of natural gas and water (research on possible usage)

Section 4.5

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Approximate Relative Fuel Consumption in the

U.S.Fuels for Electrical

Generation

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Alternate and Renewable Energy Sources

• Alternate energy sources – energy sources not based on fossil fuels or nuclear processes.

• Renewable energy sources – energy sources that cannot be exhausted.

• In large part these types of energies overlap.

Section 4.6

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Chapter 4 - Important Equations

• W = Fd (Work)

• Ek = ½mv2 (Kinetic Energy)

• Ep = mgh (Potential Energy)

• Total Energy = Ek + Ep

• P = W/t (Power)

Review

2gh• v =

Quantity Unit SymbolEquivalent Units

Force newton N Kgꞏm/s2

Work joule J Nꞏm

Energy joule J Nꞏm

Power watt W J/s

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𝑚 11,000 

𝑐 1100 

𝑘 1,000 

𝑣ΔxΔ𝑡 

𝑎Δv⃑Δ𝑡 

𝑑12 𝑔𝑡2 

𝐹 𝑚𝑎 𝑊 𝑚𝑔 𝐸 ℎ𝑓  

𝑊𝑏𝑜𝑦 𝑣𝑏𝑜𝑦 𝑊𝑔𝑖𝑟𝑙 𝑣𝑔𝑖𝑟𝑙(momentum) 

𝑊 𝐹𝑑 

𝐾𝐸 12 𝑚𝑣2 

𝑃𝐸 𝑚𝑎𝑑 

𝑣 2𝑔ℎ⃑ 

𝑃𝑊𝑡 

𝑇𝑘 𝑇𝐶 273 𝐻𝑒𝑎𝑡 𝑆𝐻 𝑚 Δ𝑡   

 

𝐻𝑒𝑎𝑡 𝐿𝑓 𝑚 

𝐻𝑒𝑎𝑡 𝐿𝑣𝑚 𝑃1

𝑃2

𝑇1

𝑇2 

𝑣 𝑓𝜆 

𝑇1𝑓 

𝑛𝑐

𝑐𝑚 

𝑉 𝐼𝑅 𝑃 𝑉𝐼  

Stuff you now know…Stuff you need for the Final

•38 •39

•40 •41

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3 | 42http://themalloryfamily.net

Bring for the final:• Pen (or pencil)• Calculator• 3”x5” card• Ruler• Test 6 (astronomy)

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Questions…4. How much work is required to lift a 4.0 kg concrete block to a

height of 2.0 m?

h = 0.0mm = 4.0 kg

h = 2.0mm = 4.0 kg

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Questions…12. Which has more kinetic energy: a 0.0020 kg bullet traveling at 400

m/s or a 6.4 x 107kg ocean liner traveling at 10 m/s (20 knots)? Justify your answer.

v = 10 m/sm = 6.4 x 107 kg

v = 400 m/sm = 0.0020 kg

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Questions…15. What is the potential energy of a 3.00 kg object at the bottom of a

well 10.0 m deep as measured from ground level? Explain the sign of the answer.

h = 0.00 mm = 3.00 kg

h = -10.0 mm = 3.00 kg

•42 •43

•44 •45

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Questions…16. How much work is required to lift a 3.00 kg object from the

bottom of a 10.0 m deep well?

h = 0.00 mm = 3.00 kg

h = 10.0 mm = 3.00 kg

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Questions…20. A 35.0 kg child starting from rest slides down a water slide with a

vertical height of 20.0 m. What is the child’s speed ata. Halfway down the slide’s vertical distance? b. Three quarters of the way down?

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Questions…20. A 35.0 kg child starting from rest slides down a water slide with a

vertical height of 20.0 m. What is the child’s speed ata. Halfway down the slide’s vertical distance? b. Three quarters of the way down?

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Questions…20. A 35.0 kg child starting from rest slides down a water slide with a

vertical height of 20.0 m. What is the child’s speed ata. Halfway down the slide’s vertical distance? b. Three quarters of the way down? h = 20.0 m

m = 35.0 kg

h = 10.0 mm = 35.0 kg

h = 5.0 mm = 35.0 kg

h = 0.0 mm = 35.0 kg

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•48 •49

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Questions…24. A 130 lb student races up stairs with a vertical height of 8.0 m in

5.0 s to get to a class on the second floor. How much power in watts does the student expend in doing work against gravity?

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Questions…24. A 130 lb student races up stairs with a vertical height of 8.0 m in

5.0 s to get to a class on the second floor. How much power in watts does the student expend in doing work against gravity?

h = 0 mw = 130 lbst = 0 s

h = 8.0 mw = 130 lbst = 5.0 s

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Don’t forget your homework!!!Test Next Week on

Chapters 3 & 4

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