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Chapter 8A. WorkChapter 8A. WorkA PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

Physics and WorkPhysics and Work

In this module, we will learn a measurable definition of work as the

product of force and distance.

Objectives: After completing this Objectives: After completing this module, you should be able to:module, you should be able to:•• Describe Describe work work in terms of force and in terms of force and

displacement, using the definition of the displacement, using the definition of the scalar productscalar product..

•• Solve problems involving concept of work.Solve problems involving concept of work.

•• Distinguish between the Distinguish between the resultantresultant work and work and the work of a single force.the work of a single force.

•• Define the Define the spring constantspring constant and calculate and calculate the work done by a varying spring force.the work done by a varying spring force.

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 F.There must be an applied force F.

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

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

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

F

W

The force The force F F exerted on the exerted on the pot by the man does work.pot by the man does work.

The earth exerts a force The earth exerts a force W W on on pot, but does no work even pot, but does no work even though there is displacement.though there is displacement.

Definition of WorkDefinition of Work

Work is a scalar quantity equal to the product of the displacement x and the component of the force Fx in the direction of the displacement.

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

Work = Force component X displacementWork = Force component X displacement

Work = Fx xWork = 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, then, then

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

Work = 160 JWork = 160 J 1 1 NNmm = = 1 Joule (J)1 Joule (J)

Negative WorkNegative Work

fx

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

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

Work = Work = ((--10 N)(4 m) = 10 N)(4 m) = -- 40 J40 J

Work = - 40 JWork = - 40 J

Resultant work is the algebraic sum of Resultant work is the algebraic sum of the individual works of each force.the individual works of each force.

Example:Example: F = F = 4040 N, f = N, f = --10 N and 10 N and x = x = 4 m4 m

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

Work = 120 JWork = 120 J

Resultant Work or Net WorkResultant Work or Net Work

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)


40 N4 m -10 N

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

Work of a Force at an AngleWork of a Force at an Angle

x = x = 1212 mmF = F = 70 N70 N

6060ooWork = Fx x

Work = (F cos ) x

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

Work = 420 JWork = 420 JOnly the xOnly the x--component of component of

the force does work!the force does 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 formula.3. Find work of a single force from formula.4. Resultant work is work of resultant force.4. Resultant work is work of resultant 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 ground. with the ground. What is the work of this force?What is the work of this force?

300

x = 20 m

F = 200 N

Work = Work = (F (F coscos ) x) x

Work = Work = (200 N)(20 m) Cos 30(200 N)(20 m) Cos 3000

Work = 3460 JWork = 3460 J

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

FF

Example 2:Example 2: A A 4040--NN force pulls a force pulls a 44--kgkg block a block a horizontal distance of horizontal distance of 8 m8 m. The rope makes an . The rope makes an angle of angle of 353500 with the floor and with the floor and uukk = 0.2= 0.2. What is . What is the work done by each acting on block?the work done by each acting on block?

1. Draw sketch and 1. Draw sketch and find given valuesfind given values.

x P

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

2. Draw free2. Draw free--body body diagram showing diagram showing all forces. (Cont.)all forces. (Cont.)

Work = (F cos ) x+x

40 N40 N

xx

n

mg8 m

P

fk

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

+x

40 N40 N

xx

n

W = mg

8 m

P

fk

Work = (P Work = (P coscos ) x) x


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

WorkWork PP = (40 N) = (40 N) coscos 353500 (8 m) = (8 m) = 262 J262 J

5. Next consider normal force 5. Next consider normal force n n and weight and weight WW..Each makes a 90Each makes a 9000 angle angle with with xx, so that the works , so that the works are zero. (are zero. (coscos 909000=0):=0):

WorkWork PP == 0 0 WorkWork nn == 00

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

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

+x

40 N40 N

xx

n

W = mg

8 m

P

fk


WorkP = 262 J

Workn = WorkW = 0

Recall: Recall: ff kk = = kk nnnn + P + P coscos 353500 –– mg = 0;mg = 0; nn = mg = mg –– P P coscos 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

ff kk = = kk n n = = (0.2)(16.3 N);(0.2)(16.3 N); fk = 3.25 N

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

+x

40 N40 N

xx

n

W = mg

8 m

P

fkWorkP = 262 J

WorkWorknn = WorkW = 0

6. Work of friction (Cont.) 6. Work of friction (Cont.)

fk = 3.25 N; x = 8 m

Workf = (3.25 N) cos 1800 (8 m) = -26.0 J

Note work of friction is Note work of friction is negative negative coscos 18018000 = = --117. The resultant work is the sum of all works:7. The resultant work is the sum of all works:

262 J + 0 + 0 – 26 J (Work)R = 236 J(Work)R = 236 J

Example 3:Example 3: What is the resultant work on a What is the resultant work on a 44--kgkg block sliding from top to bottom of the block sliding from top to bottom of the 303000 inclined plane? (inclined plane? (hh = 20 m= 20 m and and kk = 0.2= 0.2))

Work = (F Work = (F coscos ) x) xh

300

nf

mg

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

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

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

hx300 0

20 m 40 msin 30

x 0sin 30 hx

Example 3(Cont.):Example 3(Cont.): What is the resultant work What is the resultant work on on 44--kgkg block? (block? (hh = 20 m= 20 m and and kk = 0.2= 0.2))

hh

303000

nnff

mgmg

x = x = 4040 mm1. First find 1. First find work of mg.work of mg.

Work = Work = (4 kg)(9.8 m/s(4 kg)(9.8 m/s22)(40 m) Cos 60)(40 m) Cos 6000

Work = 784 JWork = 784 JPositive Positive WorkWork

Work = Work = mg(mg(coscos ) x) x

600

mg

x2. Draw free2. Draw free--

body body diagramdiagram

Work done by weight mg

mg mg coscos

Example 3 (Cont.):Example 3 (Cont.): What is the resultant work What is the resultant work on on 44--kgkg block? (block? (hh = 20 m= 20 m and and kk = 0.2= 0.2))

hh

303000

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. Free4. Free--body diagram:body diagram:nn

ff

mgmg

mgmg coscos 303000

303000

nn = mg = mg coscos 303000= (4)(9.8)(0.866)= (4)(9.8)(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


5. Find work of friction force f 5. Find work of friction force f using freeusing free--body diagrambody diagram

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

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

Work = Work = (272 J)((272 J)(--1) = 1) = --272 J272 J

Note: Work of friction is Note: Work of friction is NegativeNegative..

ff

xx

18018000

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

hh

303000

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.


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 J272 J

Resultant Work = 512 JResultant Work = 512 J

hh

303000

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. DisplacementGraph of Force vs. DisplacementAssume that a constant force F acts Assume that a constant force F acts through a parallel displacement through a parallel displacement xx..

Force, FForce, F

Displacement, xDisplacement, x

FF

xx 11 xx 22

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(x 22 -- xx 11 ))

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(x 22 -- xx 11 ))

Work F x 40 N40 N

Force, FForce, F

Displacement, xDisplacement, x

1 m1 m 4 m4 m

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


Work of a Varying ForceWork of a Varying Force

Our definition of work applies only for a Our definition of work applies only for a constantconstant force or an force or an averageaverage force.force.

What if the force What if the force variesvaries with displacement as with displacement as with stretching a spring or rubber band?with stretching a spring or rubber band?

FF

xx FF

xx

HookeHooke’’s Laws LawWhen a spring is stretched, there is a When a spring is stretched, there is a restoringrestoring force that is proportional to the displacement.force that is proportional to the displacement.

F = -kxF = -kx

The spring constant The spring constant kk is a property is a property of the spring given by:of the spring given by:

K = F

x

FF

xx

m

Work Done in Stretching a SpringWork Done in Stretching a Spring

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 HookeFrom Hooke’’s law: F = s law: F = kxkx

x

F

Work = Area of Triangle Work = Area of Triangle

Area = Area = ½½ (base)(height) (base)(height) = = ½½ ((x)(Fx)(F avgavg ) = ) = ½½ x(kxx(kx))

Work = ½ kx2Work = ½ kx2

Compressing or Stretching a Spring Compressing or Stretching a Spring Initially at Rest:Initially at Rest:Two forces are Two forces are always present: always present: the outside force the outside force FF outout ON ON spring and spring and the reaction force the reaction force FF ss BY BY the spring.the 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

mCompressing

Stretching

Example 4:Example 4: A A 44--kgkg mass suspended from a mass suspended from a spring produces a displacement of spring produces a displacement of 20 cm20 cm. . What is the spring constant? What is the spring constant?

FF20 cm20 cm

m

The stretching force is the weight The stretching force is the weight (W = mg) of the (W = mg) of the 44--kgkg mass:mass:

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

Now, from HookeNow, from Hooke’’s law, the force s law, the force constant k of the spring is:constant k of the spring is:

k = = F

x

0.2 mk = 196 N/mk = 196 N/m

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

Work = Work = ½½(196 N/m)(0.30 m)(196 N/m)(0.30 m)22

Work = 8.82 JWork = 8.82 J

212Work kx

F

30 cm

Note: The work to stretch Note: The work to stretch an an additionaladditional 30 cm30 cm is is greater due to a greater greater due to a greater average force.average force.

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

2 21 12 12 2Work kx kx

x1 x2

Fx1

mx2

m

SummarySummary

xxFF

6060ooWork = Work = FF xx xx

Work = (F Work = (F coscos ) x) x

Work is a scalar quantity equal to the product of the displacement x and the component of the force Fx in the direction of the displacement.

WorkWork is a is a scalar quantityscalar quantity equal to the equal to the product of the displacement product of the displacement xx and the and the component of the force component of the force FF xx in the in the direction of the displacement.direction of the 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 formula.3. Find work of a single force from formula.4. Resultant work is work of resultant force.4. Resultant work is work of resultant force.

1. Always draw a free1. Always draw a free--body diagram, body diagram, choosing the positive xchoosing the positive x--axis in the axis in the same direction as the displacement.same direction as the displacement.

Important Points for Work ProblemsImportant Points for Work Problems::

2. Work is negative if a component of the 2. Work is negative if a component of the force is opposite displacement directionforce is opposite displacement direction

3. Work done by any force that is at right 3. Work done by any force that is at right angles with displacement will be zero (0).angles with displacement will be zero (0).

4. For resultant work, you can add the works 4. For resultant work, you can add the works of each force, or multiply the resultant of each force, or multiply the resultant force times the net displacement.force times the net displacement.

Summary For SpringsSummary For Springs

FF

xx

m

HookeHooke’’s Law:s Law:

F = F = --kxkxSpring Spring

Constant:Constant: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

mx2

m

Springs: Positive/Negative WorkSprings: Positive/Negative Work

x

mx

m

+

CompressingStretching

Two forces are Two forces are always present: always present: the outside force the outside force FF outout ON ON spring and spring and the reaction force the reaction force FF s s BY BY the spring.the 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 Chapter 8A -- WorkWork

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