Physics 231 Lecture 10 • • •

Main points of today’s lecture: Cons. of Energy with Gravity Potential energy of spring

PE = •

1 2 kx 2

Work, W k energy and d nonconservative and external forces

KE f − KE 0 + PE f − PE 0 = W nonconservative + Wext •

Power P = W ; P = Fv ave ave ave

Δt

Conceptual problem •

Two marbles, T bl one ttwice i as h heavy as th the other, th are dropped d d ((nott thrown) to the ground from the roof of a building. Just before hitting the ground, the heavier marble has – a)) as much h ki kinetic ti energy as th the lilighter ht one. – b) twice as much kinetic energy as the lighter one. – c) half as much kinetic energy as the lighter one. – d) four times as much kinetic energy as the lighter one. – e) impossible to determine M1 = m;

M 2 = 2m

Dropped from rest KE i = 0

Conservation of energy for mass 1 KE f ,1 − KE i,1 = mg(yi − y f )

Dropped from rest

Conservation of energy for mass 2

KE f ,1 = mg(yi − yf )

KE f ,2 − KE i,2 = 2mg(y ( i − yf )

KE f ,2 = 2mg(yi − yf )

clicker question •

A 0.400-kg 0 400 kg bead slides on a curved wire wire, starting from rest at point A in the figure below. If the wire is frictionless, find the speed of the bead at C.

A C

B

y0

5m

yf

2m

v0

0

vf

?

KE f + PE f = KE 0 + PE 0

KE 0 = 0 – – – –

a) 5.2 m/s b) 1.4 m/s c) 23 m/s d) 7.7 m/s

KE f = PE 0 − PE f = mg(y0 − yf ) 1 mvf2 = mg(y0 − yf ) 2  vf = 2g(y0 − yf ) = 7.7m / s

Work and PE for non-constant forces •

When the Wh th force f is i nott constant, t t the th workk can still till be b computed t d for f small displacements Δx over which force is approximately constant:

W = Fx Δx •

To get the total work, we add the contributions for the steps:

W =  Fx Δx

•

W is just the area under the curve, if the force is conservative the change in potential energy is –W. It is equal to the work one would need to do against the force to move the object over the chosen path.

Work and potential energy for a spring • •

The force is a spring is not constant: Fx ,spring = −kx; k is the spring constant Let’s move the mass attached below by hand from x=0 to x with constant velocity. This means Fhand+Fspring=0, Fhand= -Fspring=kx Fhand=kx

x=0

•

x

x

The PE is the area under this curve. It is a triangle: A=1/2•base•height

PE =

1 (kx )x = 1 kx 2 2 2

Example

•

x0

0.1m

xf

0

v0

0

k

2 N/m

m

0 5 kg 0.5

vf

?

The spring above is stretched from it equilibrium length x=0 to a maximum i length l h off xmax=0.1 0 1 m, then h it i is i released. l d The h spring i constant is k=2 N/m. What is the speed of a 0.5kg mass when it returns to the equilibrium length? 1 2 x f = 0  PE f = 0 PE kx 0 ; = 0 -2 2 – a)) 2x10 2 10 m/s / 2 1 -1 KE = 0; KE f = mvf2 – b) 2x10 m/s 0 2 – c) 1x10-3 m/s KE 0 + PE 0 = KE f + PE f -2 – d) 1x10 m/s

1 2 1 kx 0 = mvf2 2 2 k vf2 = x 02  v f = m



k x 0 = 0.2m / s m

Conceptual quiz •

A spring-loaded i l d d ttoy d dartt gun iis used d tto shoot h tad dartt straight t i ht up in the air, and the dart reaches a maximum height of 24 m.The same dart is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the dart go this time, neglecting friction and assuming an ideal spring? KE + PE f f ,gravity + PE f ,spring – a) 48 m = KE 0 + PE 0,gravity + PE 0,spring – b) 24 m KE f = KE 0 = 0; PE 0,gravity = PE f ,spring = 0 – c) 12 m – d) 6 m  PE f ,gravity = PE 0,spring – e) impossible to determine Call x to be the initial displacement of the mass on the spring

vo

0

vf

0

y0

0

h1,f

24 m

mgh1,f

h2,f

?

mgh 2,f

mgh f =

1 2 kx 2

1 2 2 2 k ( x1 )   x 1   = 2  h 2,f =  2  h1,f =   ⋅ 24m 1 2  2  x1  k ( x2 ) 2

Work-energy theorem with both conservative, nonconservative and external forces Wnon − cons + Wext = E mech,f − E mech,0

 In I an isolated i l t d system, t with ith conservative ti forces, f mechanical h i l energy is conserved! •

Proof: The work-energy work energy theorem:

W = Wcons + Wnon − cons + Wext = KE f − KE 0 •

For conservative forces, Wcons= - ΔPE:

− ( PE f − PE 0 ) + Wnon − cons + Wext = KE f − KE 0  Wnon − cons + Wext = KE f − KE 0 + PE f − PE 0 E mech = KE + PE  Wnon − cons + Wext = E mech,f − E mech,0

Example A skier starts from rest at the top of a hill that is inclined at 20° with the horizontal The hillside is 200 m long horizontal. long, and the coefficient of friction between snow and skis is 0.075. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. What is the distance d that the skier glides g the horizontal p portion of the snow before coming g to rest? along

KE f + PE f − KE 0 − PE 0 = Wnon − cons . KE 0 = 0; KE f = 0.

PE f − PE 0 = Wnon − cons .

h

PE f − PE 0 = mg(yf − y0 ) = − mg(200m) sin(20o ). Wnon − cons = Wf ,hill + Wf ,flat Wf ,flat = − μ k mgd

Wf ,hill = −f k (200m) = − μ k N(200m) = − μ k mg cos(20o )(200m)  − mg(200m) sin(20o ) = − μ k mg // / / cos(20o )(200m) − μk mgd //  sin(20o ) .342 . o  d = (200m)  − cos(20 )  = 200m( − .940) = 724m. 0.075  μk 

Forms of Energy Mechanical Energy

Ug

K

Thermal Energy

Us

Other forms include

Eth

Echem

Enuclear Slide 10-12

Some Energy Transformations

Echem → Ug

K → Eth

Echem → Ug

Us → K → Ug

Slide 10-15

Energy Transformations conserve energy Kinetic energy K = energy of motion Potential energy U = energy of position Thermal energy Eth = energy associated i t d with ith temperature t t System energy E = K + U + Eth + Echem + ... Energy is a conserved property of an isolated system. y Energy can be transformed within the system without loss or gain of the energy of the isolated system. Slide 10-14

Power •

P Power iis the th rate t att which hi h workk is i being b i done. d Th average power is The i

Pav = •

W Δt

If we chose the x-axis to lie along the force and displacement is characterized by an average velocity: v

W Fx Δx Pav = = = Fx v Δt Δt •

The SI units for power are J/s = W (Watts). Another unit is hp (horsepower). 1 hp = 746 W.