Chapter 7 Energy and Chapter 7 Energy and Energy TransferEnergy Transfer

Introduction to Introduction to EnergyEnergy

The concept of energyenergy is one of the most important topics in science

Every physical process that occurs in the Universe involves energy energy and energy transfersenergy transfers or transformationstransformations

Energy is not easily defined!!Energy is not easily defined!!

Energy Approach to Energy Approach to ProblemsProblems

The energy approach to describing motion is particularly useful when the force is not constant

An approach will involve Conservation Conservation of Energyof Energy This could be extended to biological

organisms, technological systems and engineering situations

7.1 Systems and 7.1 Systems and EnvironmentsEnvironments

A systemsystem is a small portion of the Universe We will ignore the details of the rest of

the Universe A critical skill is to identifyto identify the system A valid systemvalid system may

be a single object or particle be a collection of objects or particles be a region of space vary in size and shape

Problem SolvingProblem Solving Does the problem require the

system approach? What is the particular system and

what is its nature? Can the problem be solved by the

particle approach? The particle approach is what we

have been using to this time

EnvironmentEnvironment There is a system boundarysystem boundary around

the system The boundary is an imaginary surface It does not necessarily correspond to a

physical boundary The boundary divides the system from

the environmentenvironment The environment is the rest of the

Universe

7.2 Work Done by a 7.2 Work Done by a Constant ForceConstant Force

The work, work, WW, done on a system by an agent exerting a constant force on the system is the product of the magnitude, magnitude, FF,, of the force, the the magnitude magnitude rr of the displacement of the point of application of the force, and cos cos where is the angle between the force and the displacement vectors

Work, cont.Work, cont. WW = = FF rr cos cos (7.1)(7.1)

The displacement is that of the point of application of the force

A force does no workdoes no work on the object if the force does not move through a displacement

The work The work done by a force on a moving objectdone by a force on a moving object is is zero zero when the force appliedwhen the force applied is perpendicular to is perpendicular to the displacement the displacement of its point of applicationof its point of application

Work is a scalar quantityWork is a scalar quantity The unitunit of work is a joule (J)joule (J)

1 joule = 1 NEWTON 1 joule = 1 NEWTON 1 meter 1 meter J = N J = N ·· m m

Work ExampleWork Example The normal force, nnormal force, n,

and the gravitational gravitational force, force, mmg,g, do no do no workwork on the object cos = cos 90° = 0

The force FF does dodoes do work on the object

Example 7.1 Example 7.1 ConceptualConceptual ExampleExample

If the magnitude of FF is held constant but the angle θθ is increased, What happened with the work WW done by FF?

DECREASES!!!DECREASES!!! Since: cosθ

DecreasesDecreases when 0 < 0 < θθ < 90 < 90oo

Δr

Can exert a force & do Can exert a force & do no work!no work! Example:Example: Walking at

constant vv with a grocery bag.

W = W = FFrr coscosθθ Could have rr = 0, F = 0, F ≠≠ 0 0

W = 0W = 0Could have F F rr θθ = 90º, cos= 90º, cosθθ = 0 = 0

W = 0W = 0

Example 7.2 Example 7.2 ConceptualConceptual ExampleExample

ΔΔrr

More About WorkMore About Work The systemsystem and the environment environment must

be determined when dealing with work The environment does work ondoes work on the system Work byby the environment on on the system

The signsign of the work depends on the direction of FF relative to rr Work is positivepositive when projectionprojection of FF onto

DDrr is in the same directionsame direction as the displacement

Work is negativnegative when the projection is in the opposite directionopposite direction

More About Work, 2More About Work, 2 Cause of the DisplacementCause of the Displacement

A force is not not necessarily the cause of the object’s displacement.

Example:Example: Lifting an objectLifting an object.. Work is done byby Gravitational Force (FFGG), although FFGG is not the cause of the object moving upward!upward!

More About Work, More About Work, finalfinal

Displacement with Displacement with FrictionlessFrictionless An object is displaced on a

frictionless horizontal surface, the normal force nn and the gravitational force mmgg do no work on the object.

Example:Example: In the situation shown here, the force applied FFPP is the only force doing work onon the object.

Work Is An Energy Work Is An Energy TransferTransfer

This is important for a system approach to solving a problem

If the workwork is done on a system and it is positivepositive, energy is transferred tois transferred to the system

If the workwork done on the system is negativenegative, energy is transferred fromis transferred from the system

If a system interactsinteracts with its environment, this interaction can be described as a transfer of transfer of energyenergy across the system boundary This will result in a changechange in the amount of amount of

energy storedenergy stored in the system

Example 7.3 Example 7.3 Work Done Work Done ONON a Box a Box

Given:Given: m = 50 kg, Fm = 50 kg, FPP = 100N, F = 100N, Ffrfr = 50N, = 50N, ΔΔr = r = 40.0m40.0m

Find:Find: (a). Work done BYBY each force. (b). Net work Done ONON the box.

Example 7.3 Example 7.3 Work Done Work Done ONON a Box, 2a Box, 2

For each force ONON the box: WW = F = Frr cos cosθθ (a)(a)

WWG G == mg mgΔΔrrcos90cos90oo = mg(40m)(0) = = mg(40m)(0) = 00

WWn n == n nΔΔrrcos90cos90oo = n(40m)(0) = = n(40m)(0) = 00

WWP P == F FPPΔΔrrcos37cos37oo = (100N)(40m)(cos37 = (100N)(40m)(cos37oo) =) = 3200J3200J

WWfrfr== F FfrfrΔΔrrcos180cos180oo = (50N(40m)( = (50N(40m)(––1)=1)= –– 2000J 2000J(b)(b) WWnetnet == WWGG++ WWn n + + WWPP+ + WWfrfr = = 1200J1200J WWnetnet == (F(Fnetnet))xx ΔΔr = (r = (FFPPcos37cos37oo – F – Ffrfr) ) ΔΔr r

WWnetnet == (100Ncos37(100Ncos37oo – 50N)40m – 50N)40m == 1200J 1200J

Active Figure 7.4Active Figure 7.4

Example 7.4 Example 7.4 Work Done Work Done ONON a Backpacka Backpack

Find The net worknet work done ONON the backpack.

From (a): h = dh = dcoscosθθ From (b): ΣΣFy = 0 Fy = 0

FFHH = = mmgg From (c):

Work done BYBY the the hikerhiker::

WWHH == FFHH ddcoscosθθ

WWHH == mmghgh

Example 7.4 Example 7.4 Work Done Work Done ONON a a Backpack, 2Backpack, 2

From (c): Work done BYBY gravitygravity::

WWGG = = mmgg ddcoscos(180 (180 – – θθ))

WWGG = = mmgg d(d(––coscosθθ) = ) = – – mmgg ddcoscosθθ

WWGG = =–– mmghgh NET WORKNET WORK on the backpack: on the backpack:

WWnetnet = = WWHH ++ WWGG == mghmgh – – mgh mgh

WWnetnet = 0= 0

FFGG exerted BYBY the Earth ONON the Moon acts toward the Earth and provides its centripetal accelerationcentripetal acceleration inward along the radius orbit

FFGG r r (Tangent to the circle & parallel with

v)The angle The angle θθ = 90 = 90oo

WWE-ME-M = = FFGG r r coscos 9090o o = 0= 0

This is why the Moon, as well as This is why the Moon, as well as artificial satellites, can stay in orbit artificial satellites, can stay in orbit without expenditure ofwithout expenditure of FUEL!!!FUEL!!!

Example 7.5 Example 7.5 ConceptualConceptual Example:Example: Work Work BY BY the Earth the Earth ONON the Moon. the Moon.

7.3 Scalar Product of Two 7.3 Scalar Product of Two VectorsVectors

The scalar product of two vectors is written as A A . . BB It is also called the dot productdot product

A A . . B B = = A BA B cos cos (7.2)(7.2) is the angle betweenbetween AA and BB

The scalar product is commutative A A . . B B == B B . . AA

The scalar product obeys the distributive law of multiplication A A . . (B + C) (B + C) == A A . . B + A B + A . . CC

Dot Products of Unit Dot Products of Unit VectorsVectors

(7.4) & (7.5)(7.4) & (7.5)

Using component form with AA and BB:

(7.6)(7.6)

(Problem #6)(Problem #6)

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Example 7.6 Example 7.6 Work Done by a Work Done by a Constant Force Constant Force (Example 7.3 Text Book)

Given:Given: r = r = (2.0 (2.0 î î ++ 3.03.0 ĵĵ) m ) m F F = (5.0 = (5.0 î î ++ 2.02.0 ĵĵ) N) N

Calculate the following magnitudes: rr = (4 + 9)= (4 + 9)½½ = (13) = (13)½½ = 3.6 m = 3.6 m FF = (25 + 4)= (25 + 4)½½ = (29) = (29)½½ = 5.4 N = 5.4 N

Calculate the Work done by F:F:

WW = = FF •• ΔΔr r == [[(5.0 (5.0 î î ++ 2.02.0 ĵĵ) N][(2.0 ) N][(2.0 î î ++ 3.03.0 ĵĵ) m]) m] = (5.0 (5.0 îî •• 2.0 2.0 î î + + 5.0 5.0 îî •• 3.03.0 ĵ ĵ + + 2.02.0 ĵĵ •• 2.0 2.0 îî + + 2.02.0 ĵĵ •• 3.03.0 ĵĵ)) NN •• mm

= [10 + 0 + 0 + 6] J == [10 + 0 + 0 + 6] J = 16 J16 J

Examples to Read!!!Examples to Read!!! Example 7.5Example 7.5 (page 189) Example 7.6Example 7.6 (page 193)

Homework to be solved in Homework to be solved in Class!!!Class!!! Questions: 1Questions: 1 Problems: 1, 6 Problems: 1, 6 (Done: See Eqn 7.6)(Done: See Eqn 7.6)

Material for the FinalMaterial for the Final