Energy Conservation1. Mechanical energy conservation

0 UKE

For closed isolated system

UKE

2. Open system

extWEUK int

3. Conservative and nonconservative forces

•Forces such as gravity or the elastic force, for which the work dose not depend on the path taken but only on the initial and final position, are called conservative forces

•For conservative forces the work done on a closed path (a lop) is equal to zero

•Friction is a nonconservative force

nonconsWE int

consWU

Example: A block is moved from rest at point A to rest at point B.Which path requires the most work to be done on the object?

A) The table is leveled and friction is present.Path 1 Path 2 Path 3 All the same

B) The table is tilted and frictionless. Path 1 Path 2 Path 3 All the same

123

A

B

Example: A hammer slides along 10 m down a 30 inclined roof and off into the yard, which is 7 m below the roof edge. Right before it hits the ground, its speed is 14.5 m/s. What is the coefficient of kinetic friction between the hammer and the roof?

Δx = 10 m

h = 7 m

v = 14.5 m/s

30This can be solved using Newton’s laws and kinematics, but it’s looooooooooooooooooooooooong.

21sin cos

2 kmv mg x mg x mgh

22 sin

2 cosk

g h x v

g x

22(9.8) 7 10sin30 14.51.5

2(9.8)10cos30

h’

Example: In the system below, a 200 g box is pushed 4 cm against a spring with k = 250 N/m and released. The box slides along a frictionless horizontal surface and then up an incline which makes an angle of 30 with respect to the horizontal. The coefficient of kinetic friction between the box and the incline is 0.2. How far along the incline is the box when its speed is half its maximum speed?

d ?Compression:

x = 4 cm

θ = 30

μk = 0.2m = 200 g k = 250 N/m

vMAX

vMAX/2

2initial

10

2E kx

2fi nal

10

2E mv

(MAX)

kv x

m

250 N/ m(0.04 cm)

0.2 kg

1.4 m/ s

initial fi nalE E

1) In the first part of the motion, mechanical energy is conserved.

2) For the whole process, mechanical energy E = K + Ug + Uelastic is not conserved due to friction: ΔE = Wfriction

2initial

10 0

2E kx

kf rictionW f d

2

MAXfi nal

10

2 2

vE m mgh

2

2MAX1 12 2 2 k

vm mgh kx f d

k coskf mg (MAX)

kv x

m

2 21 1sin cos

8 2 kkx mgd kx mgd

23

8 sin cosk

kxd

mg

2

2

3(250 N/ m)(0.04 cm)8(0.2 kg)(9.8 m/ s ) sin30 0.2cos30

0.114 m 11.4 cm

sindh

1D)

2

Examples:

12

y

x

dUU mgy F mg

dy

dUU kx F kx

dx

The force is minus the slope of the

U (x) curve

Relation between U and F (conservative force)

final

initial

xdxFWU

tconsdxFU x dx

dUFx

x

U

xx = 0, F = 0

dU/dx = 0

x

U

xx < 0, F > 0

dU/dx < 0

221 kxxU

Spring

kxxF

x

U

xx > 0, F < 0

dU/dx > 0

x

U

xx > 0, F < 0 and larger in

magnitude

dU/dx > 0 and steeper

than before

The force always points “downhill”!!!

A

Example: Which of the force versus position graphs matches the potential energy function U(x)?

U

x

F

x

x

xF

F

B

C

+ −Slope: 0

0

0

Force = − slope !

1D)

Relation between U and F (conservative force)

2 and 3D)

, , in cartesian coordinates

f or the radial component in spherical coordinates

x y z

r

U U UF F F

x y z

UF

r

UF The force is minus

the gradient of the function U (x,y,z)

constrdrFU

)(

The force is minus the slope of the

U (x) curve tconsdxFU x

dx

dUFx

Visualization of a gradient in 2D

Think of a hilly terrain where U is the altitude.

The negative gradient of U is a vector whose:• Direction points down the hill in the direction water would flow from

that location (i.e., in the steepest direction). • Magnitude is the slope of the hill in that direction

x

y

U

x

y

U

U

Example: Find the force exerted at point P (0,1,2) m if the potential energy associated with the force is:

Pˆˆ ˆ3 8 12F i j k

2 3( ) 3 4U r xy x yz

,P3 8 (3 0) 3x x

UF y x F

x

3,P3 (0 8) 8y y

UF x z F

y

2,P3 ( 3 1 4) 12z z

UF yz F

z

23 3ˆ3ˆ38 yzjxziyxrF

Whenever F = 0 (ie, dU/dx = 0), we have equilibrium.

x

U

xS

xU xN

xS, xU and xN are points of equilibrium

Equilibrium

x

U

The force brings it back to the equilibrium point.

stable

The force pulls it away from the equilibrium point.unstabl

e

The force remains zero, so the particle stays at the new position, which is also an equilibrium position.

neutral

What happens if the particle moves some small dx away from the equilibrium point?

Stable/unstable/neutral equilibrium

• Minimum = stable equilibrium

• Maximum = unstable equilibrium

• Force points “downhill”

• Turning points : E =U (so K = 0)

x

U

UMIN

KEMAX

x0

212 oE kx

UMAX ( = E )

KE = 0, turn-around points

Particle moves here

Forbidden region

(KE < 0)

Forbidden region

(KE < 0)

–x0

Energy Diagrams Example 1: A box attached to a spring on a horizontal, frictionless table is released at x = x0 from rest. 2

0210 kxUKE

Example 2: The box is brought to x = x0 and pushed, so its initial velocity is v0.

2 21 12 2o oE KE U mv kx

xt–xt

New turn-around points.

x

U

x0

21

2 oE kx (before)

2 21 12 2o oE mv kx (now)

x

U

E

xt–xt

U

KE

How much kinetic/potential energy does the system have at every point?

U

KE

U = 0

KE = KEMAX = E

U = UMAX = E

KE = 0

U

x

A particle is subjected to the force associated with this potential. No other forces are exerted on the particle. Describe the motion of the particle in the following cases.

Example: Potential with two pits.

1. The particle is released from rest at point A.

U

xA

UA

At M1, U is minimum, so K (and speed) is maximum

M1

The particle oscillates between A and B.

B

At xB, U = E, so K (and speed) is zero → turn around point

Direction of force F

E

From the initial conditions, 0 AE K U E U

forbiddenforbidden OKOK

The particle is forbidden from x < xA or x > xB (K < 0)

2. The particle is released at point A with a small* initial velocity v0.

U

xA

UA

2A A0

1From the initial conditions, (*but not too much larger)

2E mv U U

E

At M1, U is minimum, so K (and speed) is maximum

M1

The turn-around points are defined by K = 0, so U = E : points C and D.

DC

The particle oscillates between C and D.

Direction of force Fforbiddenforbidden OKOK

The particle keeps moving in the +x direction (no oscillations).

Direction of force F

forbidden OK OK OK

3. The particle is released from rest at point G.

U

xG

UG

GFrom the initial conditions, E U

E

At M1, U is minimum, so K (and speed) is maximum

M1

4. The particle is released from rest at point H. The particle has maximum speed at point:

U

xH

A. M1

B. M2

C. M3

M3M1

M2 E

J

The particle oscillates between H and J.

Direction of force F

forbidden forbiddenOK

5. The particle is released from rest at point K.

U

xK

E

From initial conditions, E = UK

OK

The particle oscillates between K and L.

forbiddenforbidden

L K

6. The particle is released from rest at point M1.

U

xM1

E

From initial conditions, E = UM1

Equilibrium

Force = 0 with v = 0 →

Direction of force F

If someone pushes the particle slightly away from M1, the force pushes it back.

Stable equilibrium

7. The particle is released from rest at point M2.

U

x

M2

E

From initial conditions, E = UM2

Equilibrium Force = 0 with v = 0 →

Direction of force F

If someone pushes the particle slightly away from M1, the force pushes it further away.

Unstable equilibrium