SPH4U: Introduction to WorkSPH4U: Introduction to Work

Work & EnergyWork & Energy DiscussionDiscussion DefinitionDefinition

Dot ProductDot Product Work of a constant forceWork of a constant force

Work/kinetic energy theoremWork/kinetic energy theorem Work of multiple constant forcesWork of multiple constant forces CommentsComments

Work & EnergyWork & Energy

One of the most important concepts in physicsOne of the most important concepts in physics Alternative approach to mechanicsAlternative approach to mechanics

Many applications beyond mechanicsMany applications beyond mechanics Thermodynamics (movement of heat)Thermodynamics (movement of heat) Quantum mechanics...Quantum mechanics...

Very useful toolsVery useful tools You will learn new (sometimes much easier) ways to You will learn new (sometimes much easier) ways to

solve problemssolve problems

Forms of EnergyForms of Energy

KineticKinetic: : Energy of motionEnergy of motion.. A car on the highway has kinetic energy.A car on the highway has kinetic energy.

We have to remove this energy to stop it.We have to remove this energy to stop it. The brakes of a car get The brakes of a car get HOTHOT!! This is an example of turning one form of energy into This is an example of turning one form of energy into

another (thermal energy). another (thermal energy).

Energy ConservationEnergy Conservation Energy cannot be Energy cannot be destroyed destroyed or or created.created.

Just changed from one form to another.Just changed from one form to another.

We say We say energy is conservedenergy is conserved!! True for any True for any closedclosed system system.. i.e. when we put on the brakes, the kinetic energy of the car is turned into heat using friction in the brakes. The total energy of the “car-brakes-road-atmosphere” system is the same.i.e. when we put on the brakes, the kinetic energy of the car is turned into heat using friction in the brakes. The total energy of the “car-brakes-road-atmosphere” system is the same. The energy of the car “alone” is The energy of the car “alone” is notnot conserved... conserved...

It is reduced by the braking.It is reduced by the braking.

Doing “Doing “workwork” on an isolated system will change its “” on an isolated system will change its “energyenergy”...”...

Definition of Work:Definition of Work:

Ingredients: Ingredients: Force (FF), displacement (rr)

Work, W, of a constant force FF

acting through a displacement r r is:

W = FF rr = F rr cos = Fr rr

FF

rr

displace

ment

Fr

“Dot Product”“Dot Product”

The dot product allows us to multiply two vectors, but just the components that are going in the same direction (usually along the second vector)

Definition of Work...Definition of Work... Only the component of Only the component of FF along the displacement is along the displacement is

doing work.doing work. Example: Train on a track.Example: Train on a track.

FF

rr

F cos

Aside: Dot Product (or Scalar Aside: Dot Product (or Scalar Product)Product)

Definition:

aa.bb = ab cos

= a[b cos ] = aba

= b[a cos ] = bab

Some properties:aabb = bbaaq(aabb) = (qbb)aa = bb(qaa) (q is a scalar)aa(bb + cc) = (aabb) + (aacc) (cc is a vector)

The dot product of perpendicular vectors is 0 !!

aa

ab bb

aa

bb

ba

Work & Energy UnderstandingWork & Energy Understanding A box is pulled up a rough (A box is pulled up a rough ( > 0 > 0) incline by a rope-pulley-weight arrangement as shown below.) incline by a rope-pulley-weight arrangement as shown below.

How many forces are doing work on the box?How many forces are doing work on the box?

(a)(a) 2

(b)(b) 3

(c)(c) 4

Box

SolutionSolution

Draw FBD of box:N

f

mg

T Consider direction of

motion of the box

v

Any force not perpendicularto the motion will do work:

N does no work (perp. to v)

T does positive work

f does negative work

mg does negative work

3 forcesdo work

Work: Example Work: Example (constant force)(constant force)

A force A force FF = 10 N = 10 N pushes a box across a frictionlesspushes a box across a frictionlessfloor for a distance floor for a distance x x = 5 m= 5 m..

xx

FF

Work done byby F F onon box is:

WF = FFxx = F x (since FF is parallel to xx)

WF = (10 N) x (5 m) = 50 Joules (J)Joules (J)

Units of Work:Units of Work:

Force x Distance = WorkForce x Distance = Work

N-m (Joule) Dyne-cm (erg)

= 10-7 J

BTU = 1054 J

calorie = 4.184 J

foot-lb = 1.356 J

eV = 1.6x10-19 J

cgs othermks

Newton x

[M][L] / [T]2

Meter = Joule

[L] [M][L]2 / [T]2

Work & Kinetic Energy:Work & Kinetic Energy:

A force A force FF = 10 N = 10 N pushes a box across a frictionlesspushes a box across a frictionlessfloor for a distance floor for a distance x x = 5 m= 5 m. The speed of the box is . The speed of the box is vv11

before the push and before the push and vv22 after the push.after the push.

xx

FFv1 v2

m

Work & Kinetic Energy...Work & Kinetic Energy...

Since the force Since the force FF is constant, acceleration is constant, acceleration aa will be will be constant. We have shown that for constant constant. We have shown that for constant aa:: vv22

22 - v - v1122 = 2 = 2aa(x(x22-x-x11) = 2) = 2aaxx..

multiply by multiply by 11//22mm:: 11//22mvmv222 2 - - 11//22mvmv11

22 = m = maaxx

But But FF = = mmaa 11//22mvmv2222 - - 11//22mvmv11

22 = F = Fxx

xx

FFv1 v2

aam

Work & Kinetic Energy...Work & Kinetic Energy...

So we find thatSo we find that 11//22mvmv22

22 - - 11//22mvmv1122 = F = Fx = Wx = WFF

DefineDefine Kinetic Energy Kinetic Energy KK: : K = K = 11//22mvmv22

KK22 - K - K11 == WWFF

WWFF == K K (Work/kinetic energy theorem)(Work/kinetic energy theorem)

xx

FFaam

v2v1

Work/Kinetic Energy Theorem:Work/Kinetic Energy Theorem:

{{NetNet WorkWork done on object}done on object}

==

{{changechange inin kinetic energykinetic energy of object}of object}

netW K

2 1K K

2 22 1

1 1

2 2mv mv

University will prove this for a variable force later.

Work & Energy QuestionWork & Energy Question

Two blocks have masses Two blocks have masses mm11 and and mm22, where , where mm11 > > mm22. They are sliding on a frictionless floor and have the . They are sliding on a frictionless floor and have the same same

kinetic energykinetic energy when they encounter a long rough stretch (i.e. when they encounter a long rough stretch (i.e. > 0 > 0) which slows them down to a stop.) which slows them down to a stop.Which one will go farther before stopping?Which one will go farther before stopping?

(a)(a) m1 (b)(b) m2 (c)(c) they will go the same distance

m1

m2

SolutionSolution

The work-energy theorem says that for any object The work-energy theorem says that for any object WWNETNET = = KK In this example the only force that does work is In this example the only force that does work is ffriction (since both riction (since both NN and and mgmg are perpendicular to the block’s motion). are perpendicular to the block’s motion).

mf

N

mg

SolutionSolution

The work-energy theorem says that for any object The work-energy theorem says that for any object WWNETNET = = KK In this example the only force that does work is In this example the only force that does work is ffriction (since riction (since

both both NN and and mgmg are perpendicular to the blocks motion). are perpendicular to the blocks motion). The net work done to stop the box is - The net work done to stop the box is - fD = fD = --mgD.mgD.

m

D

This work “removes” the kinetic energy that the box had: WNET = K2 - K1 = 0 - K1

SolutionSolution

The net work done to stop a box is The net work done to stop a box is - - fD = fD = --mgDmgD.. This work “removes” the kinetic energy that the box had:This work “removes” the kinetic energy that the box had: WWNETNET = K = K22 - K - K11 = 0 - K = 0 - K11

This is the same for both boxes (same starting kinetic energy).This is the same for both boxes (same starting kinetic energy).

m2gD2m1gD1 m2D2m1D1

m1

D1

m2

D2

Since m1 > m2 we can see that D2 > D1

A Simple Application:A Simple Application:Work done by gravity on a falling objectWork done by gravity on a falling object What is the speed of an object after falling a distance What is the speed of an object after falling a distance HH, assuming it , assuming it

starts at rest?starts at rest? WWgg = = FF r r = = mgmg rr cos cos((00) ) = = mgHmgH

WWgg = = mgmgHH

Work/Kinetic Energy Theorem:Work/Kinetic Energy Theorem:

WWg g = mgH= mgH = = 11//22mvmv22

rrmgg

H

jj

v0 = 0

v 2v gH

What about multiple forces?What about multiple forces?

Suppose FFNET = FF1 + FF2 and the

displacement is rr.

The work done by each force is:

W1 = FF1 r r W2 = FF2 rr

WTOT = W1 + W2

= FF1 r r + FF2 rr

= (FF1 + FF2 ) rr

WTOT = FFTOT rr It’s the totaltotal force that matters!!

FFNET

rrFF1

FF2

Comments:Comments:

Time interval not relevantTime interval not relevant Run up the stairs quickly or slowly...same Run up the stairs quickly or slowly...same WorkWork

SinceSince WW = = FF rr

No work No work is done if:is done if: FF = = 0 0 oror rr == 0 0 oror = = 9090oo

Comments...Comments...

WW == FF rr

No work done if No work done if = 90= 90oo..

No work done by No work done by TT..

No work done by No work done by NN..

TT

v v

vvNN

Work & Energy QuestionWork & Energy Question

An inclined plane is accelerating with constant acceleration An inclined plane is accelerating with constant acceleration aa. A box resting on the plane is . A box resting on the plane is held in place by static friction. How many forces are doing work on the block?held in place by static friction. How many forces are doing work on the block?

aa

(a)(a) 1 (b)(b) 2 (c) (c) 33

SolutionSolution

First, draw all the forces in the system:First, draw all the forces in the system:

aa

mgmg NN

FFSS

aa

mgmg NN

SolutionSolution

Recall that Recall that WW = = FF Δr Δr so only forces that have a component along so only forces that have a component along the direction of the displacement are doing work.the direction of the displacement are doing work.

FFSS

The answer is (b) 2.

FlashFlash