Physics 2101 Physics 2101 Section 3Section 3

Mar. 1Mar. 1stst: Ch. 7: Ch. 7--9 review9 reviewCh. 10 Ch. 10

Announcements:Announcements:Announcements:Announcements:•• Test#2 (Ch. 7Test#2 (Ch. 7--9) will be at 9) will be at 6 PM March 3 (6) Lockett)6 PM March 3 (6) Lockett)6 PM, March 3 (6) Lockett)6 PM, March 3 (6) Lockett)

St d i M d iSt d i M d i•• Study session Monday eveningStudy session Monday eveningat 6:00PM at Nicholson 130 at 6:00PM at Nicholson 130

Class Website: http://www.phys.lsu.edu/classes/spring2010/phys2101http://www.phys.lsu.edu/classes/spring2010/phys2101--3/3/Or go directly to my websiteOr go directly to my website

Work and EnergyWork and EnergyWorkWork energy theoremenergy theorem

netW K= Δ

WorkWork--energy theoremenergy theorem

System

net

net netW F ds= ⋅∫System21

2K mv=

2

Work doneWork done E K U+Mechanical energyMechanical energy

Work done Work done by nonby non--conservative forcesconservative forces

E K UE K U≡ +

Δ = Δ +ΔWork done Work done

by conservative forcesby conservative forces .conserW U= −Δ

2. .

1( ) ; ( )2grav elasU y mgy U x kx= =Potential energyPotential energy

Work and EnergyWork and Energy

Conservation of mechanical theoremConservation of mechanical theorem

SystemConst.mecE K U= + =

SystemmecE 0Δ =

More than mechanical energyMore than mechanical energy

mec themal internalE E EEΔ ≡ Δ +Δ +Δmec themal internal

System W E= ΔSystem W E= Δ

Work and EnergyWork and Energy

0W E= Δ =

IsolatedIsolated system

Conservation of total energy:Conservation of total energy:

The total energy of an isolated system cannot changeThe total energy of an isolated system cannot change

Impulse and MomentumImpulse and Momentum

( )f i

n n

J P P P

P mv P m v

= Δ = −

= =∑

ft

t

J Fdt= ∫

Systemit

Any external impulse Any external impulse increases the momentumincreases the momentumincreases the momentum increases the momentum of a system of a system

dPF madt

= =Nothing but Newton 2Nothing but Newton 2ndnd--lawlawdt

ft

f iJ Fdt P P= = −∫i

f it

J Fdt P P∫

Impulse and MomentumImpulse and Momentum

Impulse from external forcesImpulse from external forces J∑Impulse from external forces Impulse from external forces .extJ∑

Impulse fromImpulse frominternal forcesinternal forces int 0ernalJ =∑ (collision forces)

0F∑Conservation of total Conservation of total

. 0extF =∑ Momentum:Momentum:

IsolatedSystem

The total momentum of an The total momentum of an isolated system cannot isolated system cannot changechangeSystem changechange 0n n

nP m vΔ = Δ =∑ ∑

Impulse and MomentumImpulse and Momentum

. 0extF =∑ Application to collision:Application to collision:

0P m vΔ = Δ =∑ ∑ 0n nn

P m vΔ Δ∑ ∑

Elastic collision:Elastic collision: 0n nn

P m vΔ = Δ =∑ ∑

21( ) 02 n n

nK m vΔ = Δ =∑ ∑

The total momentum and total kinetic energy cannot changeThe total momentum and total kinetic energy cannot change

Friction Force ProblemsFriction Force Problems

E K U K U Ef f f i i iE K U K U E= + ≠ + =

NonNon--conservative Forces: Frictionconservative Forces: FrictionM h i lMechanical energy:Conservative System mechE K U= +

Friction takes Energy out of the system—treat it like thermal energy

ΔE ≠ 0 L M h i lΔEmech ≠ 0 Lose Mechanical energy

I l th k d b t l f iIn general the work done by an exteW= E=

rnal force is

mech ThermalE EΔ Δ + Δ

If there is no work done by an external forceΔEmech + ΔEThermal = 0Emech( final) = Emech(initial)− Eth( final)+ Eth(initial)

ΔEThermal

mech( f ) mech( ) th( f ) th( )K f +U f = Ki +Ui − Ff • displacement

A new way to look at Friction

Question #8: A block of mass m slides down the inclined plane starting with zero velocity. Region D has friction and it comes to rest after moving a distance D.

Mechanical Energy is not Conserved.E ( final) E (initial) ΔE

We can draw this.Emec( final) = Emec(initial)− ΔEth

Emec( final) = Emec(initial)− Ff xEmec( final) = Emec(initial)− ΔEth

Emec( final) = Emec(initial)− Ff x

K =mv2

2

A new way to look at Friction

Question #9: Find K and Emec as the block moves along.

Emech-Ffdistance

Kmv2

K =2

Problem: No FrictionTh ll i l d th i li i f i ti l IfThe pulley is massless and the incline is frictionless. If the blocks are released from rest with the connecting cord taut, block B accelerated downward. What is their total kinetic energy when block B has fallen a distancetotal kinetic energy when block B has fallen a distance L?

From conservation of Mechanical Energy, the change in total potential energy is equal AND opposite (i.e. negative) to the change in the total kinetic energy.

Initially the blocks are stationary, so that KEi=0This means E h = 0

We define the potential energy to be zero initially.

When Block B has fallen L m, Block A has risen [(L)sinθ]

This means Emech 0

Thus, the final total kinetic energy is:

Emec = K f +U f = Ki +Ui = 0 + 0

K f +U f =mA + mB( )v2

2− mBgL + mALgsinθ = 0 K f =

mA + mB( )v2

2= mBgL − mALgsinθ

2

Problem: FrictionProblem: Friction

The pulley is massless and the incline has a kinetic coefficient of friction μk. If the blocks are released from rest with the connecting cord taut, block B accelerated downward. What is their total kinetic energy when block B has fallen a distance L?

Not a conservative force so Mechanical Energy is not conserved.

Initially the blocks are stationary, so that Ki=0

We define the potential energy to be zero initially This means E h = 0We define the potential energy to be zero initially.

When Block B has fallen L m, Block A has risen [(L)sinθ]: Uf = -mBgL+mAgLsinθ

This means Emech 0

The Energy removed (thermal) is the work done by friction = mAgLcosθ

K f +U f = Ki +Ui − ΔEth = −ΔEthf U f i Ui th th

K f +U f =mA + mB( )v2

2− mBgL + mALgsinθ = −mAgLμk cosθ

2Lg m m sinθ + μ cosθ( )( )K f =

mA + mB( )v2

2= mBgL − mALg sinθ + μk cosθ( ) v ==

2Lg mB − mA sinθ + μk cosθ( )( )mA + mB( )

A 10 kg block is released from point A in the figure . The track is frictionless except for the portion between points B and C, which has a length of 6 m. The block travels down the track, hits a spring of force constant 250 N/m, and compresses the spring 0.3 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of Kinetic friction between the block and the rough surface between B and C.

Lets draw the picture again!

Emec(initial) = mghEnergy lost=Ff d = mgμkdEnergy lost Ff d = mgμkd

After the block passes the friction surfaceAfter the block passes the friction surface

Emech = K +U =mv2

2= mgh − mgμkd

So when it compresses the spring v=0kx2

h d 22

= mgh − mgμkd

A block slides along a track from one level to

Problem: FrictionProblem: FrictionA block slides along a track from one level to a higher level, by moving through anintermediate valley. The track is frictionless until the block reaches the higher level There is a frictional forceblock reaches the higher level. There is a frictional force that stops the block in the distance d. The block’s initial speed is v0; the height difference is h; the coefficient of kinetic friction is μk. What is d?

What do we know? Along the part of the track which is frictionless, conservation of Mechanical Energy holds. However, at the top the friction transfers energy out of the system (ΔE h ) Since the system is “isolated”

[ ]0 W K U E= = Δ +Δ +Δ

energy out of the system (ΔEtherm). Since the system is isolated the change in the total energy is zero.

[ ],0 ext net net thermW K U E= = Δ +Δ +Δ

0 = 0− 12 mv0

2( )+ mgh −0( )[ ]+ μ k mg( )( )d

d =1

μkmg12 mv0

2 −mgh( )2 h

Solving for d

=v0

2

2μ kg−

hμk

Problem 8-63 A particle can slide along a track as shown. The curved portions of the track are frictionless, but the flat part has a coefficient of kinetic friction of . The particle is μk = 0.20released from rest at point A, which is a height h=L/2 above the flat part of the track.

Where does the particle finally stop?

12mec gravE U mgh mgL= = =

Initial mechanical energy

2mec grav g g

Energy lost to friction every time going through the flat partf kW mg Lμ= − ⋅fric kW mg Lμ

Assume the particle moves n trips (not run-trip)

1fricE W n W

L L

Δ = = ⋅ 0.5mgL

21 2 5

kmgL n mg L

n

μ− = ⋅− ⋅

= =0

2.52 k

nμ

= =Stop at the middle of the flat part!Stop at the middle of the flat part!