Chapter 8A. Chapter 8A. WorkWork

A PowerPoint Presentation byA PowerPoint Presentation byPaul E. Tippens, Professor of Paul E. Tippens, Professor of

PhysicsPhysicsSouthern Polytechnic State Southern Polytechnic State

UniversityUniversity© 2007

Three things are necessary for the Three things are necessary for the performance of work:performance of work:

F

F

x

• There must be an applied force There must be an applied force F.F.

• There must be a displacement There must be a displacement x.x.

• The force must have a The force must have a component along the component along the displacement.displacement.

If a force does not affect If a force does not affect displacement, it does no displacement, it does no

work.work.

F

W

The force The force F F exerted on exerted on the pot by the man does the pot by the man does work.work.The earth exerts a force The earth exerts a force W W on pot, but does no work on pot, but does no work even though there is even though there is displacement.displacement.

Definition of WorkDefinition of WorkWorkWork is a is a scalar quantityscalar quantity equal to equal to the product of the displacement the product of the displacement xx and the component of the force and the component of the force FFxx in the direction of the in the direction of the displacement.displacement.

Work = Force component X displacement

Work = Fx x

Positive WorkPositive WorkFx

Force F contributes to displacement x.

Example: If F = Example: If F = 4040 N and N and x = x = 4 m4 m, , thenthen

WorkWork = (40 N)(4 m) = 160 Nm

Work = 160 J 1 N1 Nmm = = 1 Joule 1 Joule (J)(J)

Negative WorkNegative Workfx

The friction force The friction force f f opposes the opposes the displacement.displacement.

Example: If f = -Example: If f = -1010 N and N and x = x = 44 mm, , thenthen

Work = Work = (-10 N)(4 m) = - 40 J(-10 N)(4 m) = - 40 J

Work = - 40 J

Resultant work is the algebraic Resultant work is the algebraic sum of the individual works of sum of the individual works of

each force.each force.

Example:Example: F = F = 4040 N, f = -10 N and N, f = -10 N and x = x = 4 4 mmWork = Work = (40 N)(4 m) + (-10 N)(4 (40 N)(4 m) + (-10 N)(4 m)m)

Work = 120 J

Resultant Work or Net Resultant Work or Net WorkWork

Fx f

Resultant work is also equal to the work of the RESULTANT

force.

Example:Example: Work = (F - f) xWork = (40 - 10 N)(4 m)

Work = 120 J

40 N4 m -10 N

Resultant Work (Cont.)Resultant Work (Cont.)

Work of a Force at an Work of a Force at an AngleAngle

x = x = 1212 mmF = F = 70 70

NN6060ooWork = Fx x

Work = (F cos ) x

Work = Work = (70(70 N) Cos 60N) Cos 6000 (12 m) = (12 m) = 420 J420 J

Work = 420 JOnly the x-component Only the x-component

of the force does of the force does work!work!

1. Draw sketch and establish what is given and what is to be found.

Procedure for Calculating Work

2. Draw free-body diagram choosing positive x-axis along displacement.

Work = (F cos ) x+

FF xxn

mg

3. Find work of a single force from 3. Find work of a single force from formula.formula.4. Resultant work is work of resultant 4. Resultant work is work of resultant force.force.

Example 1:Example 1: A lawn mower is pushed a A lawn mower is pushed a horizontal distance of horizontal distance of 20 m20 m by a force of by a force of 200 N200 N directed at an angle of directed at an angle of 303000 with the with the ground. What is the work of this force?ground. What is the work of this force?

300

x = 20 m

F = 200 N

Work = Work = (F cos (F cos ) x) xWork = Work = (200 N)(20 m) Cos (200 N)(20 m) Cos

303000

Work = 3460 J

Note: Work is positive since Fx and x are in the same direction.

FF

Example 2:Example 2: A A 40-N40-N force pulls a force pulls a 4-kg4-kg block a block a horizontal distance of horizontal distance of 8 m8 m. The rope makes . The rope makes an angle of an angle of 353500 with the floor and with the floor and uukk = 0.2= 0.2. . What is the work done by each acting on What is the work done by each acting on block?block?1. Draw sketch 1. Draw sketch

and find given and find given valuesvalues.

x P

P = P = 40 N; 40 N; xx = 8 m, = 8 m, uukk = 0.2; = 0.2; = 35 = 3500; ; mm = 4 = 4 kgkg 2. Draw free-body 2. Draw free-body

diagram diagram showing all showing all forces. (Cont.)forces. (Cont.)Work = (F cos )

x+x

40 40 NN

xx

n

mg8 m

P

fk

Example 2 (Cont.):Example 2 (Cont.): Find Work Done by Each Find Work Done by Each Force.Force.

+x

40 40 NN

xx

n

W = mg8 m

P

fk

Work = (P cos Work = (P cos ) ) xx

P = P = 40 N; 40 N; xx = 8 m, = 8 m, uukk = = 0.2; 0.2; = 35 = 3500; ; mm = 4 kg = 4 kg

4. First find work of 4. First find work of P.P.

WorkWorkPP = (40 N) cos 35= (40 N) cos 3500 (8 m) = (8 m) = 262 J262 J5. Next consider normal force 5. Next consider normal force n n and and

weight weight WW..Each makes a 90Each makes a 9000 angle with angle with xx, so that , so that the works are zero. the works are zero. (cos 90(cos 9000=0):=0):

WorkWorkPP = = 0 0 WorkWorknn = = 0 0

Example 2 (Cont.):Example 2 (Cont.):

6. Next find work of 6. Next find work of friction. friction.

+x

40 40 NN

xx

n

W = mg8 m

P

fk

P = P = 40 N; 40 N; xx = 8 m, = 8 m, uukk = = 0.2; 0.2; = 35 = 3500; ; mm = 4 kg = 4 kg

WorkP = 262 JWorkn = WorkW = 0

Recall: fRecall: fkk = = kk nnnn + P cos + P cos 353500 – mg = – mg = 0;0;

nn = mg – P cos = mg – P cos 353500

nn = = (4 kg)(9.8 m/s(4 kg)(9.8 m/s22) – (40 N)sin 35) – (40 N)sin 350 0 = = 16.3 N16.3 N

ffkk = = kk n n = = (0.2)(16.3 (0.2)(16.3 N);N);

fk = 3.25 N

Example 2 (Cont.):Example 2 (Cont.):

+x

40 40 NN

xx

n

W = mg8 m

P

fkWorkP = 262 J

WorkWorknn = WorkW = 0

6. Work of friction 6. Work of friction (Cont.) (Cont.) fk = 3.25 N; x = 8 m

Workf = (3.25 N) cos 1800 (8 m) = -26.0 JNote work of friction is Note work of friction is negative negative cos cos

18018000 = -1 = -17. The resultant work is the sum of all 7. The resultant work is the sum of all works:works:

262 J + 0 + 0 – 26 J

(Work)R = 236 J

Example 3:Example 3: What is the resultant work What is the resultant work on a on a 4-kg4-kg block sliding from top to block sliding from top to bottom of the bottom of the 303000 inclined plane? ( inclined plane? (hh = = 20 m20 m and and kk = 0.2 = 0.2))

Work = (F Work = (F cos cos ) ) xx

h300

nf

mg

xNet work = Net work = (works)(works)

Find the work of 3 forces.Find the work of 3 forces.

First find magnitude of x from First find magnitude of x from trigonometry:trigonometry:

h x300 0

20 m 40 msin 30

x 0sin 30 hx

Example 3(Cont.):Example 3(Cont.): What is the resultant What is the resultant work on work on 4-kg4-kg block? ( block? (hh = 20 m = 20 m and and kk = = 0.20.2))

hh303000

nnff

mgmg

x = x = 4040 mm

1. First find 1. First find work of work of mg.mg.

Work = Work = (4 kg)(9.8 m/s(4 kg)(9.8 m/s22)(40 m) Cos )(40 m) Cos 606000

Work = 784 J Positive Positive WorkWork

Work = mg(Work = mg(coscos ) ) xx

600mg

x2. Draw 2. Draw

free-body free-body diagramdiagram

Work done by weight

mg

mg cos mg cos

Example 3 (Cont.):Example 3 (Cont.): What is the resultant What is the resultant work on work on 4-kg4-kg block? ( block? (hh = 20 m = 20 m and and kk = = 0.20.2))

hh303000

nnff

mgmg

rr 3. Next find work of 3. Next find work of friction force friction force ff which which requires us to find requires us to find nn..

4. Free-body diagram:4. Free-body diagram:nn

ff

mgmg

mgmg cos 30 cos 3000

303000

nn = mg = mg cos 30cos 3000= (4)(9.8)= (4)(9.8)(0.866)(0.866)

nn = 33.9= 33.9 NN f = f = kk nn

f = f = (0.2)(33.9 N) = 6.79 N(0.2)(33.9 N) = 6.79 N

Example 3 (Cont.):Example 3 (Cont.): What is the resultant What is the resultant work on work on 4-kg4-kg block? ( block? (hh = 20 m = 20 m and and kk = = 0.20.2)) 5. Find work of friction 5. Find work of friction

force f using free-body force f using free-body diagramdiagram

Work Work = (6.79 N)(20 m)(cos = (6.79 N)(20 m)(cos 18018000))

Work = (f Work = (f coscos ) x) x

Work = Work = (272 J)(-1) = -272 J(272 J)(-1) = -272 JNote: Work of friction is Note: Work of friction is

NegativeNegative..

ff

xx

18018000

What work is done by the normal What work is done by the normal force force nn??

hh303000

nnff

mgmg

rr

Work of Work of nn is is 0 0 since it is at right angles to x.since it is at right angles to x.

Example 3 (Cont.):Example 3 (Cont.): What is the resultant What is the resultant work on work on 4-kg4-kg block? ( block? (hh = 20 m = 20 m and and kk = = 0.20.2))

Net work = Net work = (works)(works)

Weight: Weight: Work = Work = + 784 J+ 784 J

Force Force nn: : Work = Work = 0 J0 J Friction: Friction: Work = -Work = - 272 J 272 J

Resultant Work = 512 J

hh303000

nnff

mgmg

rr

Note: Resultant work could have been found by multiplying the resultant force by the net displacement down the plane.

Graph of Force vs. Graph of Force vs. DisplacementDisplacement

Assume that a constant force F acts Assume that a constant force F acts through a parallel displacement through a parallel displacement x.x.

Force, FForce, F

Displacement, xDisplacement, x

FF

xx11 xx22

The The area area under the under the curve is equal to the curve is equal to the work done.work done.

Work = F(xWork = F(x22 - x - x11))

Work F x

Area

Example for Constant ForceExample for Constant ForceWhat work is done by a constant force of What work is done by a constant force of 40 N 40 N moving a block from x = moving a block from x = 1 m1 m to x = to x =

4 m4 m??

Work = F(xWork = F(x22 - x - x11))

Work F x 40 N40 N

Force, FForce, F

Displacement, xDisplacement, x1 m1 m 4 4

mm

AreaWork = Work = (40 N)(4 m - 1 (40 N)(4 m - 1

m)m)Work = 120 J

Work of a Varying ForceWork of a Varying ForceOur definition of work applies only for Our definition of work applies only for a a constantconstant force or an force or an averageaverage force. force.

What if the force What if the force variesvaries with with displacement as with stretching a displacement as with stretching a

spring or rubber band?spring or rubber band?

FF

xx FF

xx

Hooke’s LawHooke’s LawWhen a spring is stretched, there is a When a spring is stretched, there is a

restoringrestoring force that is proportional to the force that is proportional to the displacement.displacement.

F = -kxThe spring constant The spring constant kk is a is a property of the spring given property of the spring given by:by:

K = F

x

FF

xx

m

Work Done in Stretching a Work Done in Stretching a SpringSpring

F

x

m

Work done Work done ONON the spring is the spring is positivepositive; work ; work BYBY the spring is the spring is

negative.negative.From Hooke’s law: F = kxFrom Hooke’s law: F = kx

x

F

Work = Area of Triangle Work = Area of Triangle Area = Area = ½½ (base)(height) (base)(height)

= = ½ ½ (x)(F(x)(Favgavg ) = ) = ½ ½ x(kx)x(kx)Work = ½ kx2

Compressing or Stretching a Compressing or Stretching a Spring Initially at Rest:Spring Initially at Rest:Two forces are Two forces are always present: always present: the outside force the outside force FFoutout ON ON spring spring and the reaction and the reaction force force FFss BY BY the the spring.spring.Compression: Fout does positive work and

Fs does negative work (see figure).Stretching: Fout does positive work and Fs does negative work (see figure).

x

mx

mCompressin

g Stretching

Example 4:Example 4: A A 4-kg4-kg mass suspended mass suspended from a spring produces a displacement from a spring produces a displacement of of 20 cm20 cm. What is the spring constant? . What is the spring constant?

FF20 cm20 cmm

The stretching force is the The stretching force is the weight (W = mg) of the weight (W = mg) of the 4-kg4-kg

mass:mass:F = F = (4 kg)(9.8 m/s(4 kg)(9.8 m/s22) = 39.2 ) = 39.2

NNNow, from Hooke’s law, the Now, from Hooke’s law, the

force constant k of the spring force constant k of the spring is:is:

k = =

F

x

0.2 mk = 196

N/m

Example 5:Example 5: What work is What work is required to stretch this spring (required to stretch this spring (k k = 196 N/m= 196 N/m) from ) from xx == 0 0 to to xx = 30 = 30 cmcm? ?

Work = Work = ½(196 N/m)(0.30 ½(196 N/m)(0.30 m)m)22

Work = 8.82 J

212Work kx

F30 cm

Note: The work to Note: The work to stretch an stretch an additionaladditional 30 30 cmcm is greater due to a is greater due to a greater average force.greater average force.

General Case for Springs:General Case for Springs:If the initial displacement is not zero, If the initial displacement is not zero, the work done is given by:the work done is given by:

2 21 12 12 2Work kx kx

x1 x2

Fx1

m

x2

m

SummarySummary

xxFF

6060ooWork = FWork = Fx x xx

Work = (F Work = (F coscos ) ) xx

WorkWork is a is a scalar quantityscalar quantity equal to equal to the product of the displacement the product of the displacement xx and the component of the force and the component of the force FFxx in the direction of the in the direction of the displacement.displacement.

1. Draw sketch and establish what is given and what is to be found.

Procedure for Calculating Work

2. Draw free-body diagram choosing positive x-axis along displacement.

Work = (F cos ) x+

FF xxn

mg

3. Find work of a single force from 3. Find work of a single force from formula.formula.4. Resultant work is work of resultant 4. Resultant work is work of resultant force.force.

1. Always draw a free-body diagram, 1. Always draw a free-body diagram, choosing the positive x-axis in the choosing the positive x-axis in the same direction as the same direction as the displacement.displacement.

Important Points for Work Important Points for Work ProblemsProblems::

2. Work is negative if a component of 2. Work is negative if a component of the force is opposite displacement the force is opposite displacement directiondirection3. Work done by any force that is at 3. Work done by any force that is at right angles with displacement will right angles with displacement will be zero (0).be zero (0).4. For resultant work, you can add the 4. For resultant work, you can add the works of each force, or multiply the works of each force, or multiply the resultant force times the net resultant force times the net displacement.displacement.

Summary For SpringsSummary For Springs

FF

xx

m

Hooke’s Law:Hooke’s Law:

F = -kxF = -kxSpring Spring

ConstanConstant:t:

Fkx

The spring constant is the force exerted BY the spring per unit change in its displacement. The spring force always opposes displacement. This explains the negative sign in Hooke’s law.

Summary (Cont.)Summary (Cont.)

Work = ½ kx2 2 21 12 12 2Work kx kx

Work to Stretch a Spring:

x1 x2

Fx1

m

x2

m

Springs: Positive/Negative Springs: Positive/Negative WorkWork

x

mx

m

+

Compressing Stretching

Two forces are Two forces are always present: always present: the outside force the outside force FFoutout ON ON spring spring and the reaction and the reaction force Fforce Fs s BY BY the the spring.spring.Compression: Fout does positive work and

Fs does negative work (see figure).Stretching: Fout does positive work and Fs does negative work (see figure).

CONCLUSION:CONCLUSION:Chapter 8A - WorkChapter 8A - Work