Lecture 7 (YY/MM/DD) - SFU.ca · Lecture 7 Today’s(Concepts: Work(&(Kine6c(Energy...

Classical Mechanics Lecture 7

Today’s Concepts:Work & Kine6c Energy

Mechanics Lecture 7, Slide 1

Name ____________________________ Date (YY/MM/DD) ______/_________/_______

SFU [email protected] Section__________ Group_____

UNIT 10: WORK AND ENERGYApproximate Classroom Time: Three 100 minute sessions

"Knowing is not enough; we must apply. Willing is not

enough; we must do."Bruce Lee

OBJECTIVES

1. To extend the intuitive notion of work as physical effort

to a formal mathematical definition of work as a function

of force and distance.

2. To develop an understanding of the physical significance

of mathematical integration.

3. To understand the concept of power and its relationship

to work.

4. To understand the concept of kinetic energy and its rela-

tionship to work as embodied in the work-energy theorem.

5. To relate experimentally determined kinetic energy loss

to frictional loss.

6. To apply concepts involving the conservation of momen-

tum, work, and energy to the analysis of breaking boards

in Karate.

© 1990-94 Dept. of Physics and Astronomy, Dickinson College Supported by FIPSE (U.S.

Dept. of Ed.) and NSF. Modified for SFU by N. Alberding, 2005.

Karate

Will not do Session 3 of Unit 10.

It is a Karate thing.

You can do it as a “project” in Unit 14 instead of the assigned project on the pendulum.

Stuff you asked about:

WHAT IS GOING ON WITH ALL THE INTEGRALS?????? HELP :(Why is it necessary to use integrals in the equa6on for work? everything. everything is difficultEverything is fine except the work done by Gravity far from earth The slides had a lot of informa6on that was confusing and it flew by. I

didn't have 6me to understand what was going on. The dot product is ocnfusingPre`y much everything. This prelecture sort of went over my head. I'll

definitely have to watch it again. The concept of posi6ve and nega6ve work is very confusing. please

explainUh oh. Hotdog.


I hope you have enough funny quotes to fit into your slides.

Stuff you asked about:

Does work also define a change in poten6al energy? For the last checkpoint ques6on, the poten6al energy is changing as the car moves up the slope. Would this effect the quan6ty of work?

Can we go over and work out some actual problems using dot product and the equa6ons used in the prelecture next class? Not just concept ques6ons but ones that would be similar to our homework ques6ons as well.

Can we go over Ques6on 1 in the prelecture? I think I'm missing something in the ques6on or something, but it doesn't se`le well with me.


The Dot Product


~A =A

x

ı + A

y

| + A

z

k

~B =B

x

ı + B

y

| + B

z

k

ı · ı =1| · | =1

k · k =1

ı · | =0

| · k =0

k · | =0

Dot Product using Unit Vectors

TextTextetc.

~A · ~B =(A

x

ı + A

y

| + A

z

k) · (Bx

ı + B

y

| + B

z

k)

=(Ax

B

x

ı · ı + A

x

B

y

ı · | + · · · + A

z

B

y

k · j + A

z

B

z

k · k)=(A

x

B

x

⇥ 1 + A

x

B

y

⇥ 0 + · · · + A

z

B

y

⇥ 0 + A

z

B

z

⇥ 1)~A · ~B =(A

x

B

x

+ A

y

B

y

+ A

z

B

z

)

Dot Product in Rectangular Coordinates

~A · ~B =(A

x

B

x

+ A

y

B

y

+ A

z

B

z

)

=|~A||~B| cos ✓

✓ = arccos

0BBBB@~A · ~B|~A||~B|

1CCCCA

Finding the angle between vectorsIf we know the xyz components of two vectors we can find the angle between them:

therefore

|⇥A| =|⇥B| = 5

⇥A · ⇥B =3 · 4 + 4 · 3 = 24

� = arccos

0BBBB@⇥A · ⇥B|⇥A||⇥B|

1CCCCA

� = arccos

24

25

!= arccos 0.96

=0.28 rad = 16

�

~A = 3ı + 4 |~B = 4ı + 3 |

Example

Work-Kinetic Energy TheoremThe work done by the net force F as it acts on an object that moves between posi6ons r1 and r2 is equal to the change in the object’s kine6c energy:


W =

~r2Z

~r1

~F · d~l

W =

~r2Z

~r1

~F · d~l = ~F · ~d0BBBBBBBBB@

~r2Z

~r1

d~l = ~d

1CCCCCCCCCA

Work-Kinetic Energy Theorem: 1-D Example

If the force is constant and the direc6ons aren’t changing then this is very simple to evaluate:


In this case since cos(0)=1

This is probably what you remember from High School.(The nota6on may be confusing though.)

F d

= Fd

car

Clicker Question

A lighter car and a heavier van, each ini6ally at rest, are pushed with the same constant force F. Aier both vehicles travel a distance d, which of the following statements is true? (Ignore fric6on)


A) They will have the same velocityB) They will have the same kine6c energyC) They will have the same momentum

F d

F d

car

van

~r2Z

~r1

~F · d~l = �K

A force pushing over some distancewill change the kine6c energy.


Concept – very important

Deriva6on – not so important

θ

Work done by gravity near the Earth’s surface


mgpat

h

= m~g · d~l1 + m~g · d~l1 + · · · + m~g · d~lN

mg

dl1dy1

dx1

mg

dl1

dl2

dlN



2

mg

dl1

dl2

dlN



Δy

= m~g · d~l1 + m~g · d~l1 + · · · + m~g · d~lN

Work-Kinetic Energy Theorem

If there are several forces ac6ng then W is the work done by the net (total) force:


You can just add up the work done by each force

H

Free Fall Fric6onless incline String

CheckPoint

Three objects having the same mass begin at the same height, and all move down the same ver6cal distance H. One falls straight down, one slides down a fric6onless inclined plane, and one swings on the end of a string.


In which case does the object have the biggest net work done on it by all forces during its mo6on?

A) Free Fall B) Incline C) String D) All the same

vp

Clicker Question

Three objects having the same mass begin at the same height, and all move down the same ver6cal distance H. One falls straight down, one slides down a fric6onless inclined plane, and one swings on the end of a string. What is the rela6onship between their speeds when they reach the bo`om?

H

Free Fall Fric6onless incline Pendulum

A) vf > vi > vp B) vf > vp > vi C) vf = vp = vi


v f vi

CheckPoint / Clicker Question

Only gravity will do work: Wg = Δ K

H

Free Fall Fric6onless incline String

A) vf > vi > vp B) vf > vp > vi C) vf = vp = vi

Wg = mgH

Δ K = 1/2 mv22


CheckPoint

A car drives up a hill with constant speed. Which statement best describes the total work WTOT done on the car by all forces as it moves up the hill?

Less than 40% got this right…

A) WTOT = 0

B) WTOT > 0

C) WTOT < 0

v


v

CheckPoint

A car drives up a hill with constant speed. Which statement best describes the total work WTOT done on the car by all forces as it moves up the hill?

A) WTOT = 0

B) WTOT > 0

C) WTOT < 0

A) The change in kine6c energy is zero because the velocity is constant. By the work-‐kine6c energy theorem, we know that work must also be zero.

B) The force of fric6on between the car's wheels and the ramp exert a force up the ramp over a par6cular distance. W=Fd, so the work is posi6ve. C) gravity is downward and car moves up the hill.


v

Clicker Question

A car drives up a hill with constant speed. How does the kine6c energy of the car change as it moves up the hill?

A) It increasesB) It stays the sameC) It decreases


CheckPoint

A box sits on the horizontal bed of a moving truck. Sta6c fric6on between the box and the truck keeps the box from sliding around as the truck drives.

µS

a

The work done on the box by the sta6c fric6onal force as the truck moves a distance D is:

A) Posi6ve B) Nega6ve C) Zero

About 60% got this right…


Physics 211 Lecture 7, Slide 22

From Before

A box sits on the horizontal bed of a moving truck. Sta6c fric6on between the box and the truck keeps the box from sliding around as the truck drives.

If the truck moves with constant accelera6ng to the lei as shown, which of the following diagrams best describes the sta6c fric6onal force ac6ng on the box:

µS

a

A B C

CheckPoint


The work done on the box by the sta6c fric6onal force as the truck moves a distance D is:

A) Posi6ve B) Zero C) Nega6ve D) Depends

D

A) because the direc6on of the sta6c fric6onal force and the direc6on the box travels are all same.

B) Since the fric6on force always has opposite direc6on from the movement direc6on, the work done is nega6ve

C) Fric6on keeps the box from sliding, thus D=0, and work done is 0.

µS

a

F

Problem

184 | C H A P T E R 6 Work and Kinetic Energy

Example 6-7 Pushing a Box

You push a box up a ramp using a constant horizontal force . For each distance ofalong the ramp, the box gains of height. Find the work done by for eachthe box moves along the ramp (a) by directly computing the scalar product from the

components of and where is the displacement, (b) by multiplying the product of themagnitudes of and by where is the angle between the direction of and the di-rection of (c) by finding (the component of in the direction of ) and multiplying itby (the magnitude of ), and (d) by finding (the component of in the direction of )and multiplying it by the magnitude of the force.

PICTURE Draw a sketch of the box in its initial and final positions. Place coordinate axes onthe sketch with the x axis horizontal. Express the force and displacement vectors in compo-nent form and take the scalar product. Then find the component of the force in the directionof the displacement, and vice versa.

SOLVE

FS

!S

!||!S

!!S

FS

F||!S

,FS

fcosf,!S

FS

!S

!S

,FS

5.00 mFS

3.00 m5.00 mFS

100-N

CHECK The component of in the direction of is This answer verifies our Part (b) result.

PRACTICE PROBLEM 6-6 (a) Find for and (b) Find A, B, and the angle between and for these vectors.

WORK IN SCALAR-PRODUCT NOTATIONIn scalar-product notation, the work dW done by a force on a particle over aninfinitesimal displacement is

6-18

INCREMENTAL WORK

where is the magnitude of and is the component of in the direction of The work done on the particle as it moves from point 1 to point 2 is

6-19

THE DEFINITION OF WORK

(If the force remains constant, the work can be expressed where is thedisplacement. In Chapter 3, the displacement is denoted and

are different symbols for the same quantity.)When several forces act on a particle whose displacement is the total work

done on it is

6-20dWtotal ! FS

1# d!S " F

S

2# d!S " Á ! (F

S

1 " FS

2 " Á ) # d!S ! (©FS

i ) # d!Sd!

S,F

S

i

¢rS!S

¢rS ! ¢x in " ¢y jn;!S

W ! FS # !

S,

W ! !2

1FS # d!S

d!S

.FS

F||d!S

d!

dW ! F||d! ! F cosfd! ! FS # d!S

d!S

FS

BS

AS

BS

! (2.0in " 8.0jn)m.AS

! (3.0in " 4.0jn)mAS # B

S

A cosf ! (213 m) cos71° ! 1.2 m.BS

AS

(a) 1. Draw a sketch of the situation (Figure 6-19).

2. Express and in component form and takethe scalar product:

!S

FS

4.00 # 102 J!

! (100 N)(4.00 m) " 0(3.00 m)W ! FS # !

S! Fx¢x " Fy¢y

!S

! (4.00in " 3.00jn)m

FS

! (100in " 0jn)N

y

4.00 m

5.00 m

θF

O

x

3.00 m

F I G U R E 6 - 1 9






SOLVE

FS

!S

!||!S

!!S

FS

F||!S

,FS

fcosf,!S

FS

!S

!S

,FS

5.00 mFS

3.00 m5.00 mFS

100-N




6-18

INCREMENTAL WORK


6-19




done on it is

6-20dWtotal ! FS

1# d!S " F

S

2# d!S " Á ! (F

S

1 " FS

2 " Á ) # d!S ! (©FS

i ) # d!Sd!

S,F

S

i

¢rS!S


W ! FS # !

S,

W ! !2

1FS # d!S

d!S

.FS

F||d!S

d!


d!S

FS

BS

AS

BS

! (2.0in " 8.0jn)m.AS

! (3.0in " 4.0jn)mAS # B

S

A cosf ! (213 m) cos71° ! 1.2 m.BS

AS



!S

FS

4.00 # 102 J!

! (100 N)(4.00 m) " 0(3.00 m)W ! FS # !

S! Fx¢x " Fy¢y

!S

! (4.00in " 3.00jn)m

FS

! (100in " 0jn)N

y

4.00 m

5.00 m

θF

O

x

3.00 m

F I G U R E 6 - 1 9






SOLVE

FS

!S

!||!S

!!S

FS

F||!S

,FS

fcosf,!S

FS

!S

!S

,FS

5.00 mFS

3.00 m5.00 mFS

100-N




6-18

INCREMENTAL WORK


6-19




done on it is

6-20dWtotal ! FS

1# d!S " F

S

2# d!S " Á ! (F

S

1 " FS

2 " Á ) # d!S ! (©FS

i ) # d!Sd!

S,F

S

i

¢rS!S


W ! FS # !

S,

W ! !2

1FS # d!S

d!S

.FS

F||d!S

d!


d!S

FS

BS

AS

BS

! (2.0in " 8.0jn)m.AS

! (3.0in " 4.0jn)mAS # B

S

A cosf ! (213 m) cos71° ! 1.2 m.BS

AS



!S

FS

4.00 # 102 J!

! (100 N)(4.00 m) " 0(3.00 m)W ! FS # !

S! Fx¢x " Fy¢y

!S

! (4.00in " 3.00jn)m

FS

! (100in " 0jn)N

y

4.00 m

5.00 m

θF

O

x

3.00 m

F I G U R E 6 - 1 9






SOLVE

FS

!S

!||!S

!!S

FS

F||!S

,FS

fcosf,!S

FS

!S

!S

,FS

5.00 mFS

3.00 m5.00 mFS

100-N




6-18

INCREMENTAL WORK


6-19




done on it is

6-20dWtotal ! FS

1# d!S " F

S

2# d!S " Á ! (F

S

1 " FS

2 " Á ) # d!S ! (©FS

i ) # d!Sd!

S,F

S

i

¢rS!S


W ! FS # !

S,

W ! !2

1FS # d!S

d!S

.FS

F||d!S

d!


d!S

FS

BS

AS

BS

! (2.0in " 8.0jn)m.AS

! (3.0in " 4.0jn)mAS # B

S

A cosf ! (213 m) cos71° ! 1.2 m.BS

AS



!S

FS

4.00 # 102 J!

! (100 N)(4.00 m) " 0(3.00 m)W ! FS # !

S! Fx¢x " Fy¢y

!S

! (4.00in " 3.00jn)m

FS

! (100in " 0jn)N

y

4.00 m

5.00 m

θF

O

x

3.00 m

F I G U R E 6 - 1 9

Solution

Try It Yourself

The Scalar Product S E C T I O N 6 - 3 | 185

(b) Calculate where is the angle betweenthe directions of the two vectors as shown. Equatethis expression with the Part-(a) result and solvefor cos Then solve for the work:f.

fF! cosf,

so

and

4.00 ! 102 J"W " F! cosf " (100 N)(5.00 m)0.800

cosf "Fx¢x # Fy¢y

F!"

(100 N)(4.00 m) # 0(100 N)(5.00 m)

" 0.800

FS # !

S" F! cosf and F

S # !S

" Fx¢x # Fy¢y

(c) Find and multiply it by ! :F|| F|| " F cosf " (100 N)0.800 " 80.0 N

4.00 ! 102 JW " F||! " (80.0 N)(5.00 m) "

(d) Multiply F and where is the component of in the direction of F

S:

!S

!||!|| ,

4.00 ! 102 JW " F!|| " (100 N) (4.00 m) "

!|| " ! cosf " (5.00 m)0.800 " 4.00 m

CHECK The four distinct calculations give the same result for the work.

TAKING IT FURTHER For this problem, computing the work is easiest using the procedurein Part (a). For other problems, the procedure in Part (b), Part (c), or Part (d) may be the eas-iest. You need to be competent in using all four procedures. (The more problem-solving toolsyou have at your disposal, the better.)

Example 6-8 A Displaced Particle

A particle undergoes a displacement During this displacement a con-stant force acts on the particle. Find (a) the work done by the force,and (b) the component of the force in the direction of the displacement.

PICTURE The force is constant, so the work W can be found by computingCombining this with the relation we can find

the component of in the direction of the displacement.

SOLVE

Cover the column to the right and try these on your own before looking at the answers.

Steps Answers

(a) 1. Make a sketch showing and (Figure 6-20).F||!S

,FS

,

FS

FS # !

S" F||!,W " F

S # !S

" Fx¢x # Fy¢y.

FS

" (3.00in # 4.00jn)N!S

" (2.00in $ 5.00jn)m.

y

x

ˆ

ˆ ˆ

ˆ

O = 2.00 mi – 5.00 mj

F = 3.00 Ni + 4.00 Nj

F||

F I G U R E 6 - 2 0

2. Compute the work done W. $14.0 JW " FS # !

S"

(b) 1. Compute and use your result to find !.!S # !

Sso ! " 229.0 m!

S # !S

" 29.0 m2,

2. Using solve for F|| .FS # !

S" F||!, $2.60 NF|| " F

S # !S >! "

CHECK From Figure 6-20, we see that the angle between and is between 90° and 180°,so we expect both and the work to be negative. Our results are in agreement with thisexpectation.

TAKING IT FURTHER Nowhere in the wording of either the problem statement or thesolution of Example 6-8 does it say that the motion of the particle is along any particularpath. Because the force is constant, the solution depends on the net displacement butnot on the path taken. The path could be a straight-line path or a curved path (Figure 6-21)and not a word in the solution would have to be changed.

PRACTICE PROBLEM 6-7 Find the magnitude of and the angle between and !S

.FS

fFS

!S

,

F||

!S

FS

Path"

F I G U R E 6 - 2 1

Try It Yourself

The Scalar Product S E C T I O N 6 - 3 | 185

(b) Calculate where is the angle betweenthe directions of the two vectors as shown. Equatethis expression with the Part-(a) result and solvefor cos Then solve for the work:f.

fF! cosf,

so

and

4.00 ! 102 J"W " F! cosf " (100 N)(5.00 m)0.800

cosf "Fx¢x # Fy¢y

F!"

(100 N)(4.00 m) # 0(100 N)(5.00 m)

" 0.800

FS # !

S" F! cosf and F

S # !S

" Fx¢x # Fy¢y

(c) Find and multiply it by ! :F|| F|| " F cosf " (100 N)0.800 " 80.0 N

4.00 ! 102 JW " F||! " (80.0 N)(5.00 m) "

(d) Multiply F and where is the component of in the direction of F

S:

!S

!||!|| ,

4.00 ! 102 JW " F!|| " (100 N) (4.00 m) "

!|| " ! cosf " (5.00 m)0.800 " 4.00 m

CHECK The four distinct calculations give the same result for the work.

TAKING IT FURTHER For this problem, computing the work is easiest using the procedurein Part (a). For other problems, the procedure in Part (b), Part (c), or Part (d) may be the eas-iest. You need to be competent in using all four procedures. (The more problem-solving toolsyou have at your disposal, the better.)

Example 6-8 A Displaced Particle

A particle undergoes a displacement During this displacement a con-stant force acts on the particle. Find (a) the work done by the force,and (b) the component of the force in the direction of the displacement.

PICTURE The force is constant, so the work W can be found by computingCombining this with the relation we can find

the component of in the direction of the displacement.

SOLVE

Cover the column to the right and try these on your own before looking at the answers.

Steps Answers

(a) 1. Make a sketch showing and (Figure 6-20).F||!S

,FS

,

FS

FS # !

S" F||!,W " F

S # !S

" Fx¢x # Fy¢y.

FS

" (3.00in # 4.00jn)N!S

" (2.00in $ 5.00jn)m.

y

x

ˆ

ˆ ˆ

ˆ

O = 2.00 mi – 5.00 mj

F = 3.00 Ni + 4.00 Nj

F||

F I G U R E 6 - 2 0

2. Compute the work done W. $14.0 JW " FS # !

S"

(b) 1. Compute and use your result to find !.!S # !

Sso ! " 229.0 m!

S # !S

" 29.0 m2,

2. Using solve for F|| .FS # !

S" F||!, $2.60 NF|| " F

S # !S >! "

CHECK From Figure 6-20, we see that the angle between and is between 90° and 180°,so we expect both and the work to be negative. Our results are in agreement with thisexpectation.

TAKING IT FURTHER Nowhere in the wording of either the problem statement or thesolution of Example 6-8 does it say that the motion of the particle is along any particularpath. Because the force is constant, the solution depends on the net displacement butnot on the path taken. The path could be a straight-line path or a curved path (Figure 6-21)and not a word in the solution would have to be changed.

PRACTICE PROBLEM 6-7 Find the magnitude of and the angle between and !S

.FS

fFS

!S

,

F||

!S

FS

Path"

F I G U R E 6 - 2 1


Work done by a Spring


I am confused about the posi6ve work and nega6ve work and also the posi6ve and nega6ve forces for the spring problems.

In this example the spring does –ve work since F and Δx are in opposite direc6on. The axes don’t ma`er.

Use the formula to get the magnitude of the work

Use a picture to get the sign (look at direc6ons)

A box a`ached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new loca6on.

During this mo6on, the spring does:

A) Posi6ve Work B) Nega6ve Work C) Zero work


Clicker Question

D


During this mo6on, your hand does:

A) Posi6ve Work B) Nega6ve Work C) Zero work

Clicker Question


D


During this mo6on, the total work done on the box is:

A) Posi6ve B) Nega6ve C) Zero

Clicker Question


D

= GMem

1r2� 1

r1

!


W =

~r2Z

~r1

~F(~r) · d~r = �r2Z

r1

GMemr2 dr

W = GMem

1r2� 1

r1

!A) Case 1 B) Case 2 C) They are the same

In Case 1 we send an object from the surface of the earth to a height above the earth surface equal to one earth radius.

In Case 2 we start the same object a height of one earth radius above the surface of the earth and we send it infinitely far away.

In which case is the magnitude of the work done by the Earth’s gravity on the object biggest?

Clicker Question

Case 1 Case 2


W = GMem

12RE� 1

RE

!= �GMem

2RE

W = GMem

11 �

12RE

!= �GMem

2RE

Case 1 Case 2

Case 1:

2RE

RE

Same!

Clicker Question Solution


Case 2:

Lecture 7 (YY/MM/DD) - SFU.ca · Lecture 7 Today’s(Concepts: Work(&(Kine6c(Energy...

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