classical mechanics lecture 9springer/phys1500/lectures/phys...classical mechanics lecture 9 today's...

Classical Mechanics Lecture 9

Today's Concepts and Examples:a) Energy and Frictionb) Potential energy & force

Midterm 2 will be held on March 13. Covers units 4-9

Unit 9 Homework DueThursday March 12 11:30 PM.No extension!

Mechanics Lecture 9, Slide 1

Homework 8. Awesome Job!

Average = 92%


Practice Exams

Phys 1500 Exams https://utah.instructure.com/courses/320947/files- Spring 2013: http://www.physics.utah.edu/~springer/phys1500/exams/MidtermExam2.pdf- Solutions: http://www.physics.utah.edu/~springer/phys1500/exams/MidtermExam2Soln.pdf- Long Sample: https://utah.instructure.com/courses/320947/files/45779670/download?wrap=1

Phys 2210 Exams- Practice : http://www.physics.utah.edu/~woolf/2210_Jui/rev2.pdf- Spring 2015: http://www.physics.utah.edu/~woolf/2210_Jui/ex2.pdf


http://www.physics.utah.edu/%7Espringer/phys1500/exams/MidtermExam2.pdfhttp://www.physics.utah.edu/%7Espringer/phys1500/exams/MidtermExam2Soln.pdfhttps://utah.instructure.com/courses/320947/files/45779670/download?wrap=1http://www.physics.utah.edu/%7Ewoolf/2210_Jui/rev2.pdfhttp://www.physics.utah.edu/%7Ewoolf/2210_Jui/ex2.pdf

Main Points


Incline with Friction: Work-Kinetic Energy

Using Work-Kinetic Energy Theorem

Same result as using Newton’s Law


H

N1

mg

N2

mg

µmg

must be negative

m

Work by Friction :Wfriction

Checkpoint

What is the total macroscopic work done on the block by all forces during this process?

A) mgH B) –mgH C) µk mgD D) 0


D

m

H

0000

=⇒=∆

===∆

tot

f

i

tot

WKKK

WK


Force from Potential Energy:1D


Force from Potential Energy in 3-d

Gradient operator


Potential Energy vs. Force

dxxdUxF )()( −=



2

21)( kxxU −=

dxxdUxF )()( −= kx−=


Demo


dxxdUxF )()( −=


Equilibrium


Equilibrium points


Block on Incline

1cos xTxdTWtension ∆=⋅= ∫ θ

mxTv

xTWvvmK

xTWWWWW

f

netif

tensiontensionnormalfrictionnet

1

122

1

cos2

cos)(21

cos

∆=

∆==−=∆

∆==++=

θ

θ

θ


Block on Incline

θµ cosxmgxdfW kkfriction ∆−=⋅= ∫

θsinxmgxdWWgravity ∆−=⋅= ∫


Block on Incline

θµθθ

θµθθθ

θ

cossincos

cossincoscos

cos2;0

12

2221

1

1

2

mgmgTxTx

xmgxmgxTxTxTK

mxTvv

KWWWW

xTxdTW

k

k

if

frictiongravitytensionnet

tension

−−∆−

=∆

∆−∆−∆=∆−∆−=∆

∆==

∆=++=

∆=⋅= ∫


Block on Incline

?0 =⇒

Energy Conservation Problems in general

)()( tUtKUKUKE ffiimechanical +=+=+=

For systems with only conservative forces acting

0=∆ mechanicalE

Emechanical is a constant


Gravitational Potential Energy

rEr

Err

−

Mr

Mrr

−


Gravitational Potential Problems

gravitymechanical hUhmvUKE )()(21 2 +=+=

conservation of mechanical energy can be used to “easily” solve problems.

Define coordinates: where is U=0?

M

M

E

Etotal

M

MMMoon

E

EEEarth

rrmGM

rrmGMrU

rmGMrU

rmGMrU

−−

−−=

−=

−=

)(

)(

)(

0)( →−=r

mGMrU E as ∞→r

rEr

Err

−

Mr

Mrr

−

Add potential energy from each source.


Trip to the moon

)2

1(21

21

21

222

2

iE

E

E

E

Ei

E

E

Ei

Ef

f

E

E

Ei

ffii

vGMRR

RGMv

GM

RmGMmv

mGMR

RmGM

RmGMmv

UKUK

−=

−

−=

−

−=

−=−

+=+

f

Ef

f

E

Ei

ii

RmGMU

KR

mGMU

mvK

−=

=

−=

=

0

21 2


Trip to the moon

%02.0000204.0

)01659.0)(01232.0()()(

)()()(

)(

)(

==

=⇒

≈

−=

−=

−−=

−=

EE

EM

EF

FMEE

EM

E

E

FME

M

EE

EM

FME

MFM

F

EFE

RURU

RR

RdMRM

RM

RdM

RURU

RdmGMRU

RmGMRU Can ignore effect

of moon for this problem at level of precision for SmartPhysics


Trip to the moon

( )

ME

M

E

ME

m

E

Ei

fME

m

f

E

ME

m

E

Ei

ddmGMcmGMb

dmGM

RmGMmva

bdxcbadaxcxbxbdaxadxcxxdbxdax

xdxcxxdb

xx

xdc

xb

xdxd

xdc

xba

RdmGM

RmGM

dmGM

RmGMmv

=−=−=

−−=

=−−++−

=−+−−

+−=−

−+−

=

−

+−−

=−

+=

−−−=−−

2

2

2

2

21

0)(0

)()()(

)()()(

21

…or you can practice solving the quadratic equation with many terms!!!


Trip to the moon

M

m

ME

Ef

ff

EM

m

E

Ei

ii

RmGM

dmGMU

mvK

dmGM

RmGMU

mvK

−−=

=

−−=

=

→

→

2

2

21

21 Can NOT ignore

effect of moon for this problem since the rocket is AT the moon in the end !!!!

McentersME

EcentersEM

RddRdd

−=−=

→

→


Trip to the moon

−−+−=

++−−=

++−−=

−−=−−

+=+

→→

→→

→→

→→

)1(21

22221

212

21

21

2

2222

2

22

ME

Em

ME

E

EME

Em

Ei

Eif

Mi

m

MEi

E

EMi

m

Ei

Eif

M

m

ME

E

EM

m

E

Eif

M

m

ME

Ef

EM

m

E

Ei

ffii

RMRM

dR

dMRM

RvGMvv

RvGM

dvGM

dvGM

RvGMvv

RGM

dGM

dGM

RGMvv

RmGM

dmGMmv

dmGM

RmGMmv

UKUK


Trip to the moon

020608.12

0453.0

0167.0

000207.0

)1(21

2

2

=

=

=

=

−−+−=

→

→

→→

Ei

E

ME

Em

ME

E

EME

Em

ME

Em

ME

E

EME

Em

Ei

Eif

RvGM

RMRM

dR

dMRM

RMRM

dR

dMRM

RvGMvv


Trip to the moon

xx


Block on Incline 2

1cos xTxdTWtension ∆=⋅= ∫ θ

11 xmgxdfW kkfriction ∆−=⋅= ∫ µ


Block on Incline 2

mxmgxTv

xmgxTWvvmK

xmgxTWWWWWW

kf

knetif

ktensionfrictiontensionnormalfrictionnet

)cos(2

cos)(21

cos

111

11122

111

∆−∆=

∆−∆==−=∆

∆−∆=+=++=

µθ

µθ

µθ


Block on Incline 2

θµθθ

θµθθθ

θ

cossincos

cossincoscos

cos2;0

12

2221

1

1

2

mgmgTxTx

xmgxmgxTxTxTK

mxTvv

KWWWW

xTxdTW

k

k

if

frictiongravitytensionnet

tension

−−∆−

=∆

∆−∆−∆=∆−∆−=∆

∆==

∆=++=

∆=⋅= ∫


Block on Incline 2

2sin xmgxdWWgravity ∆=⋅= ∫ θ


Clicker Question A.B.C.

D.

0% 0%0%0%

Suppose the potential energy of some object U as a function of xlooks like the plot shown below.

Where is the force on the object zero?A) (a) B) (b) C) (c) D) (d)

U(x)

x

(a) (b) (c) (d) dxxdUxF )()( −=



D.

0% 0%0%0%


Where is the force on the object in the +x direction?A) To the left of (b) B) To the right of (b) C) Nowhere

U(x)

x

(a) (b) (c) (d) dxxdUxF )()( −=



D.

0% 0%0%0%


Where is the force on the object biggest in the –x direction?A) (a) B) (b) C) (c) D) (d)

U(x)

x

(a) (b) (c) (d) dxxdUxF )()( −=


Classical Mechanics �Lecture 9Homework 8. Awesome Job!Practice ExamsMain PointsMain PointsIncline with Friction: Work-Kinetic EnergyWork by Friction :Wfriction

classical mechanics lecture 9springer/phys1500/lectures/phys...classical mechanics lecture 9 today's...

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