Energy. 2 categories stored (potential) and moving (kinetic) Energy What an object uses to move. Forces transfer energy between types WORK

Energy

2 categories ­  stored (potential) and moving (kinetic)

Energy ­ What an object uses to move

Forces ­ transfer energy between types

WORK

Work is ENERGY transferred TO or FROM an object by means of a FORCE acting ON the object!

**ENERGY transferred TO the object is POSITIVE WORK** **ENERGY transferred FROM the object is NEGATIVE WORK**

ENERGY

While ENERGY changes forms­­the total amount is CONSTANT

W = F

x

d

x

W = (F cos

θ

)d

W

f

= -fd

Negative work of friction is applied in the _{opposite direction as the motion} *FORCE must be      parallel to  displacement! *Conservative forces only can use  net displacement* conservation of energy 

W = (F cos

θ

)d

W = F d

F d = Fdcos

θ

DOT product

Scalar Product of 2 Vectors

θ

d

F

Fcosθ

Only the parts of each vector that are parallel (//) 

contribute

A B = B A

Dot products (2 vectors) = scalar!!

i i = j j = k k = 1

i j = i k = j k = 0

A force applied through a displacement changes the object's velocity !!

*When F constant, a is constant

F

net

= ma

vf2 = vi2 + 2ad solve for : m vf2 - vi2 = mad 2 1 mvf2 - 1mvi2 = Fd 2 2

KE = W

WORK ENERGY THEOREM

K = K

f

 ­ K

i

 = W

∆

K of a particle = worknet done ON or BY the particle

K

f

 = K

i

 + W

the K after the 

 worknet done =  

the K before the   worknet  done 

the worknet done

+        m = 0.40 kg       vi = 8.0 m/s       d = 2_πr       vf = 6.0 m/s       r = 1.5 m a) FIND:   work (done by friction) b) FIND:   μ c) Find # of rev when vf = 0 d) Why the work done not ZERO after 1 revolution?

A particle slides on a horizontal, circular track, 1.5 m in radius. Vi = 8.0 m/s. After one revolution its speed drops to 6.0 m/s because of friction.

       m = 0.40 kg       vi = 8.0 m/s       d = 2πr       vf = 6.0 m/s       r = 1.5 m a) FIND:   work (done by friction) b) FIND:   μ c) Find # of rev when vf = 0 d) Why is the work done not ZERO?  ­­> Friction ­> non­conservative force  path dependent 

A particle slides on a horizontal, circular track, 1.5 m in radius. Vi = 8.0 m/s. After one revolution its speed drops to 6.0 m/s

because of friction.

F

x

Could we find the work needed to move 

an object graphically?

How?

F

x

xo xf

Could we find the work needed to move 

an object graphically?

F

x

xo xf

Area Under the Curve      Integrals!

W = ΣFΔx = ∫ F dx

Xf xo

Fnet varies with X --> therefore a and v also depends on x

If an applied force varies with position according to

       

 F

(x)

 = 3x

3

 ­ 5, 

  where x is in meters,

how much work is done by this force on an object 

that moves from x = 4.0m to x = 7.0m?

 

 

If an applied force varies with position according to        

 F

_(x)

 = 3x

3

 ­ 5, 

  where x is in meters, how much work is done by this force on an object  that moves from x = 4.0m to x = 7.0m?    

Springs

Hooke's Law

Negative sign signifies force is ALWAYS 

directed OPPOSITE the DISPLACEMENT

Restoring force always acts 

   toward the equilibrium!

F

s

= -Kx

Hooke's Law

day 3

x = 0

F

+

F

-X

-X

+

F

s

= -Kx

day 3 "equilibrium" F= ­kx F = 0 ­F = kx F = 0 F = ­kx amax amax amax vmax vmax day 3

Work  W

s

Positive if the block ends up closer to the 

relaxed position (x = 0) than it was initially

Negative if the block ends up farther away from x = 0

Zero if the block ends up at the same distance from x = 0

day 3 x = 0 F = 0 x   positive F negative x negative F positive x x

*Block attached 

   to spring

(a)

(c)

(b)

day 3

Work done by a spring day 3 example #2 mass = 200g = 0.200 kg K = 1.4 kN/m d = .10m 600 v = 0 .10m equilibrium

A 200 g block is pressed against a spring of force constant k = 1.4 N/m until the block compresses the spring 10cm. The spring rests at the bottom of a ramp inclined at 600 _{to the horizontal. Use the}

work-energy theorem to determine how far up the incline the block moves before it stops a)if there is no friction between the block and the ramp, and b) if μk = 0.4.

Power!

P = 

W

∆

t

average power

P = 

dW

 dt

instantaneous power

P = 

dW

 dt

=

F cos

θ

dx

dt

=

Fcos

θ

(  )

dx

dt

P = FVcos

θ

P = F V

The loaded cab of an elevator has a mass of 3.0X103 _kg  and moves 210m up the shaft in 23s at a constant speed.         At what average rate does the force from        the cable do work on the cab?

Rate of work --> Find Power!

Fnet = 0

P = FVcos

θ

= FV = mg d

t

Conservative Forces

when the work done by the force is independent of the path, depending only on X1 and X2 examples:

   gravitational    spring    electrical

Nonconservative forces

when work done by force is path dependent examples:    friction    tension    applied forces

WORK

W

net

 = 

∆

K

W

nc

 + W

c

 = 

∆

K

W

nc

 = 

∆

K + 

∆

U

WORK

W

c

 = ­

∆

U

W

nc

 = 

∆

ME

W

c

=

∆

K = -

∆

U

∆

K +

∆

U = 0

(K

f

- K

i

) + (U

f

- U

i

) = 0

The total Mechanical Energy of a system

remains constant if the only force that

does the work is a conservative force!

K

i

+ U

i

= K

f

+ U

f

Conservation of Mechanical Energy

A 2.0 kg block slides along a frictionless track from point A to B.

How much work is done if the total distance is 2.0m along the 

track and the net vertical distance is 0.80m?

A

Potential Energy!

U

g

= mgh Gravitational potential energy!

U

s

= kx

1

2

Elastic potential energy!

2

Two blocks are connected by a light string that passes over a  frictionless pulley.  The block of mass m1 lies on a horizontal surface  and is connected to a spring of force k.  The system is released from  rest when the spring is unstretched.  If the hanging block of mass m2  falls a distance h before coming to rest, calculate μk between the  block of mass m1 and the surface. m2 m1 h call this h = 0

m2

m1

h

Relationship between

Conservative Forces and Potential Energy

** remember that potential energy function is associated only with a              CONSERVATIVE force! * the change in the potential energy of a particle under the action of a  conservative force equals the negative of the work done by the force! *Wc = ­∆U

dU = ­F

_x

dx

F

x

 =

­ dU

dx

the conservative force equals the negative derivative of the potential energy with respect to x.

*

check this:

day 7

day 7

*F = - slope of the U vs X graph

U

day 7 example #2

A potential energy function for a 

TWO

 dimensional force is of the 

form:

    

       

U = 3x

3

y ­ 7x

Find the force that acts 

     at the point (x,y).

*

Express force in unit vector notation

day 7 example #2

A potential energy function for a TWO dimensional force is of the form:     

    

U = 3x

3

y ­ 7x

Find the force that acts at the point (x,y).

The potential energy of a two particle system separated  by a distance r is given by 

U(r) = A/r

, where A is a constant. Find the radial force 

F

r.

where U(r) = 

A

R

The potential energy of a two particle system separated  by a distance r is given by 

U(r) = A/r

, where A is a constant. Find the radial force 

F

r.

where U(r) = 

A

R

Do we know of any forces with      dependance?

1

r

2

Do we know of any forces with      dependance?

1

r

2

Gravity!

Electric!

PE

g

 = ­GMm

d PE

g 

 = GMm

PE

E

 = ­kq

1

q

2

d PE

E

 = kq

1

q

2

r

r

dr      r

2

= Fg

 

!!!!

dr       r

2

= FE

 

!!!!

all conservative forces can be derived 

from the 

work­energy

 theorem!!!!

day 8 review example

Find the angle between the two vectors given by           A = ­5i ­ 3j + 2k      B = ­2j ­ 2k

The center of mass of a system of particles is the point that moves as though all of the system's mass were concentrated there!

Center of mass

When you find the center of mass of an object you can consider the  entire object to be at that point What is going on? All the forces acting on each individual piece of an object either add  or cancel out and the net force (sum of all forces on each part of the  object) acts on the center of mass 

Center of mass

m

1

x

1 +

m

2

x

2 +

m

3

x

3

m

1

+ m

2

+ m

3

X

com

=

The average position of the system's mass The center of mass can be located  by its position vector rc rcm =    mi ri M

r

cm

=

mr/ m

center of mass of an object

uniform mass

distribution

*Uniform mass distribution has Uniform density

Center of mass of Solid Bodies

*Uniform Bodies

The "particles" 

become 

differential mass 

elements dm

M is the total mass

Three masses located in the xy plane have the following coordinates: 3kg mass (3,-2)

4kg mass (-2,4) 1kg mass (2,2)

Motion of a System of Particles

Position vector of the center of mass

Time derivative of the position vector (assuming mass remains constant)

** The total momentum of the system equals the total mass

multiplied by the velocity of the center of mass!

Conservation of Momentum

Total linear momentum of an isolated system

"system" includes all particles that are interacting!

** If no external forces are acting on a system---->

∆

mv always = 0

MOMENTUM IS CONSERVED IF NO EXTERNAL FORCES!

remains constant!

Momentum

A mass that is moving has momentum!

P = mv

∆

P =

∆

mv

F

net

= m

∆

v Ft =

∆

mv

t

impulse =

∆

momentum

Impulse- momentum theorem

states that

IMPULSE produced by a net force produces a

change in momentum !

**** F

net

=m Δv  =  Δp  

t      t

dp

dt

F

net

 =

dp

dt

This is how Newton 

wrote his second 

law!!!

dp = F

net

 dt

p =  F

net

 dt

Collisions!

Elastic Collisions

momentum conserved

KE conserved

Inelastic Collisions

momentum conserved

KE 

NOT

 conserved

Head­on Elastic Collisions

m

1

 = m

2 V = 0 V = 0 ** exchange speeds

m

1

v

i1

+m

2

v

i2

 = m

1

v

f1

+m

2

v

f2

 

Head­on Elastic Collisions

m

1

<m

2 V = 0

** move off in 

opposite directions

m

1

v

i1

+m

2

v

i2

 = m

1

v

f1

+m

2

v

f2

 

Head­on Elastic Collisions

m

1

>m

2 V = 0

** Both move off in the 

direction in which m

1

 

was originally traveling

m

1

v

i1

+m

2

v

i2

 = m

1

v

f1

+m

2

v

f2

 

Glancing Elastic Collision

A A A B B B

θ

φ

V = 0

When m

1

=m

2 

 and v

i2

=0, then θ + φ = 90

o

m1vix1+m2vix2 = m1vfx1+m2vfx2 

m1viy1+m2viy2 = m1vfy1+m2vfy2 

Inelastic Colisions

m

1

v

i1

+m

2

v

i2

 = (m

1

+m

2

)v

f

 

Change in mass also produces a change 

in momentum!

Rockets! 

Same exhaust speed 

(speed of material leaving the engine)

 

but expels mass!

Same force on smaller mass         THRUST

THRUST

 on a rocket is the 

FORCE

 exerted on the 

rocket by the ejected exhaust gasses

3

rd

 Law!!

THRUST

 on a rocket is the 

FORCE

 exerted on the 

rocket by the ejected exhaust gasses

** Momentum is Conserved

** Kinetic Energy is 

NOT

 Conserved

** Kinetic Energy 

INCREASES

THRUST

 on a rocket is the 

FORCE

 exerted on the 

rocket by the ejected exhaust gasses

** Momentum is Conserved

** Kinetic Energy is 

NOT

 Conserved

** Kinetic Energy 

INCREASES

Where does that energy come from?

The internal potential (chemical potential) energy of 

the fuel!!

Rocket Propulsion

M   mass of rocket including fuel Δm  mass of fuel ve    velocity of ejected fuel (constant)

F

net

 = ma

Conservation of Momentum

On the launch pad v = 0    **momentum = 0

0 = M Δv + v

e

 ΔM

where Δm = ­ΔM

and   Δv and ve are in opposite directions

M dv = ­v

e

 dM

F

net 

= dp = Thrust =  M dv = ­v

e 

dM

Fnet = dp = Thrust =  M dv = ­ve dM

dt       dt      dt Rocket Propulsion