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Physics 207: Lecture 14, Pg 1 Lecture 14Goals: Assignment: l l HW6 due Wednesday, Mar. 11 l l For...

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Physics 207: Lecture 14, Pg 1 Lecture 14 Goals: Goals: Assignment: Assignment: HW6 due Wednesday, Mar. 11 For Tuesday: Read Chapter 12, Sections 1-3, 5 & 6 do not concern yourself with the integration process in regards to Chapter 10 Chapter 10 Understand spring potential energies & use Understand spring potential energies & use energy diagrams energy diagrams Chapter 11 Chapter 11 Understand the relationship between force, displacement and work Recognize transformations between kinetic, potential, and thermal energies Define work and use the work-kinetic energy theorem Use the concept of power (i.e., energy per time)
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Physics 207: Lecture 14, Pg 1

Lecture 14Goals:Goals:

Assignment: Assignment: HW6 due Wednesday, Mar. 11 For Tuesday: Read Chapter 12, Sections 1-3, 5 & 6

do not concern yourself with the integration process in regards to “center of mass” or “moment of inertia””

• Chapter 10 Chapter 10

• Understand spring potential energies & use energy diagramsUnderstand spring potential energies & use energy diagrams

• Chapter 11Chapter 11 Understand the relationship between force, displacement and work Recognize transformations between kinetic, potential, and thermal energies Define work and use the work-kinetic energy theorem Use the concept of power (i.e., energy per time)

Physics 207: Lecture 14, Pg 2

Energy for a Hooke’s Law spring

Associate ½ kx2 with the “potential energy” of the spring

m

2212

212

212

21 ffii mvkxmvkx

xf

xi

f

i

v

vxx

x

xx dvmvdxF

)( eqxxkvdxdv

mdtdx

dxdv

mdtdv

mF xxxx

x

dxxxkdvmvdxF xxx )( eq

Physics 207: Lecture 14, Pg 3

Energy for a Hooke’s Law spring

m

2212

212

212

21 ffii mvkxmvkx

constantK UK fsfisiU

Ideal Hooke’s Law springs are conservative so the mechanical energy is constant

Physics 207: Lecture 14, Pg 4

Energy diagrams In general:

Ene

rgy

K

y

U

Emech

Ene

rgy

K

x

U

Emech

Spring/Mass systemBall falling

Physics 207: Lecture 14, Pg 5

Equilibrium

Example Spring: Fx = 0 => dU / dx = 0 for x=xeq

The spring is in equilibrium position

In general: dU / dx = 0 for ANY function establishes equilibrium

stable equilibrium unstable equilibrium

U U

Physics 207: Lecture 14, Pg 6

Comment on Energy Conservation

We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. Mechanical energy is lost:

Heat (friction)Bending of metal and deformation

Kinetic energy is not conserved by these non-conservative forces occurring during the collision !

Momentum along a specific direction is conserved when there are no external forces acting in this direction. In general, easier to satisfy conservation of momentum

than energy conservation.

Physics 207: Lecture 14, Pg 7

Mechanical Energy Potential Energy (U)

Kinetic Energy (K) If “conservative” forces

(e.g, gravity, spring) then

Emech = constant = K + U

During Uspring+K1+K2 = constant = Emech

Mechanical Energy conserved

Before

During

After

1 2

Physics 207: Lecture 14, Pg 8

Energy (with spring & gravity)

Emech = constant (only conservative forces) At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0 Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2

Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0

Given m, g, h & k, how much does the spring compress? Em1 = Em3 = mgh = -mgx + ½ kx2 Solve ½ kx2 – mgx +mgh = 0

1

32

h

0-x

mass: m

Physics 207: Lecture 14, Pg 9

Energy (with spring & gravity)

When is the child’s speed greatest?

(A) At y1 (top of jump)

(B) Between y1 & y2

(C) At y2 (child first contacts spring)

(D) Between y2 & y3

(E) At y3 (maximum spring compression)

1

32

h

0-x

mass: m

Physics 207: Lecture 14, Pg 10

Energy (with spring & gravity)

When is the child’s speed greatest? A: Calculus soln. Find v vs. spring displacement then maximize

(i.e., take derivative and then set to zero) B: Physics: As long as Fgravity > Fspring then speed is increasing

Find where Fgravity- Fspring= 0 -mg = kxVmax or xVmax = -mg / k

So mgh = Ug23 + Us23 + K23 = mg (-mg/k) + ½ k(-mg/k)2 + ½ mv2

2gh = 2(-mg2/k) + mg2/k + v2 2gh + mg2/k = vmax2

1

32

h

0-x

mg kx

Physics 207: Lecture 14, Pg 11

Inelastic Processes If non-conservative” forces (e.g, deformation, friction)

then

Emech is NOT constant

After K1+2 < Emech (before)

Accounting for this loss we introduce Thermal Energy (Eth , new)

where Esys = Emech + Eth = K + U + Eth

Before

During

After

1 2

Physics 207: Lecture 14, Pg 12

Energy & Work

Impulse (Force vs time) gives us momentum transfer Work (Force vs distance) tracks energy transfer Any process which changes the potential or kinetic energy

of a system is said to have done work W on that system

Esys = W

W can be positive or negative depending on the direction of energy transfer

Net work reflects changes in the kinetic energy

Wnet = K

This is called the “Net” Work-Kinetic Energy Theorem

Physics 207: Lecture 14, Pg 13

Circular Motion

I swing a sling shot over my head. The tension in the rope keeps the shot moving at constant speed in a circle.

How much work is done after the ball makes one full revolution?

v(A) W > 0

(B) W = 0

(C) W < 0

(D) need more info

Physics 207: Lecture 14, Pg 14

Examples of “Net” Work (Wnet)

Examples of No “Net” Work

K = Wnet

Pushing a box on a rough floor at constant speed Driving at constant speed in a horizontal circle Holding a book at constant height

This last statement reflects what we call the “system”

( Dropping a book is more complicated because it involves changes in U and K, U is transferred to K )

K = Wnet

Pushing a box on a smooth floor with a constant force; there is an increase in the kinetic energy

Physics 207: Lecture 14, Pg 15

Changes in K with a constant F

xf

xi

f

i

v

vxx

x

xx dvmvdxF

If F is constant

xf

xi

f

i

v

vxx

x

xx dvmvdxF

KmvmvxFxxF xixfxifx 2212

21)(

Physics 207: Lecture 14, Pg 16

Net Work: 1-D Example (constant force)

Net Work is F x = 10 x 5 N m = 50 J= 10 x 5 N m = 50 J 1 Nm ≡ 1 Joule and this is a unit of energy Work reflects energy transfer

x

A force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m.

F = 0° Start Finish

Physics 207: Lecture 14, Pg 17

Units:

N-m (Joule) Dyne-cm (erg)

= 10-7 J

BTU = 1054 J

calorie = 4.184 J

foot-lb = 1.356 J

eV = 1.6x10-19 J

cgs Othermks

Force x Distance = Work

Newton x

[M][L] / [T]2

Meter = Joule

[L] [M][L]2 / [T]2

Physics 207: Lecture 14, Pg 18

Net Work: 1-D 2nd Example (constant force)

Net Work is F x = -10 x 5 N m = -50 J= -10 x 5 N m = -50 J

Work reflects energy transfer

x

F

A force F = 10 N is opposite the motion of a box across a frictionless floor for a distance x = 5 m.

= 180° Start Finish

Physics 207: Lecture 14, Pg 19

Work in 3D….

2212

21)( zizfzifz mvmvzFzzF

x, y and z with constant F:

2212

21)( yiyfyify mvmvyFyyF

2212

21)( xixfxifx mvmvxFxxF

2222

2212

21

with zyx

ifzyx

vvvv

KmvmvzFyFxF

Physics 207: Lecture 14, Pg 20

Work: “2-D” Example (constant force)

(Net) Work is Fx x = F cos(= F cos(-45°) x = 50 x 0.71 Nm = 35 J = 50 x 0.71 Nm = 35 J Work reflects energy transfer

x

F

A force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m and y = 0 m

= -45°

Start Finish

Fx

Physics 207: Lecture 14, Pg 21

Useful for performing projections.

A î = Ax

î î = 1 î j = 0

î

A

Ax

Ay

A B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )

Calculation can be made in terms of components.

Calculation also in terms of magnitudes and relative angles.

Scalar Product (or Dot Product)

A B ≡ | A | | B | cos

You choose the way that works best for you!

A · B ≡ |A| |B| cos()

Physics 207: Lecture 14, Pg 22

Scalar Product (or Dot Product)

Compare:

A B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )

with A as force F, B as displacement r

and apply the Work-Kinetic Energy theorem

Notice:

F r = (Fx )(x) + (Fy )(z ) + (Fz )(z)

Fx x +Fy y + Fz z = K

So here

F r = K = Wnet

More generally a Force acting over a Distance does Work

Physics 207: Lecture 14, Pg 23

Definition of Work, The basics

Ingredients: Force ( F ), displacement ( r )

“Scalar or Dot Product”

r

displace

ment

FWork, W, of a constant force F

acts through a displacement r :

W = F · r (Work is a scalar)(Work is a scalar)

If we know the angle the force makes with the path, the dot product gives us F cos and r

If the path is curved at each point

and rdFdW

rdF

f

i

r

rrdFW

Physics 207: Lecture 14, Pg 24

Remember that a real trajectory implies forces acting on an object

Only tangential forces yield work! The distance over which FTang is applied: Work

a

vpathand time

a

a

a = 0

Two possible options:

Change in the magnitude of v

Change in the direction of v

a = 0

a = 0

aaa= +

aradialatanga= +

FradiallFtangF= +

Physics 207: Lecture 14, Pg 26

ExerciseWork in the presence of friction and non-contact forces

A. 2

B. 3

C. 4

D. 5

A box is pulled up a rough ( > 0) incline by a rope-pulley-weight arrangement as shown below. How many forces (including non-contact ones) are doing work on the box ? Of these which are positive and which are negative? Use a Free Body Diagram Compare force and path

v

Physics 207: Lecture 14, Pg 28

Lecture 14

Assignment: Assignment: HW6 due Wednesday, March 11HW6 due Wednesday, March 11 For Tuesday: Read Chapter 12, Sections 1-3, 5 & 6For Tuesday: Read Chapter 12, Sections 1-3, 5 & 6

do do notnot concern yourself with the integration process concern yourself with the integration process


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