PH 221‐3A Fall 2009

Review for test 2Lecture 13Lecture 13

Chapters 5‐8 

(Halliday/Resnick/Walker, Fundamentals of Physics 8th edition)

The potential Energy Curve

If we plot the potential energy U versus x for aIf we plot the potential energy U versus x for a force F that acts along the x‐axis we can get a wealth of information about the motion of a particle on which F is acting.  The first parameter p g pthat we can determine is the force F(x) using the equation:

( )( ) dU xF xdx

= −

An example is given in the figures below.                                                                        In In fig.a we plot U(x) versus x.                                                                                           In fig.b we plot F(x) versus x.g p ( )

For example at x2 , x3 and x4 the slope of the  U(x) vs x   curve is zero, thus F = 0.  

The slope dU/dx between x3 and x4 is negative;  Thus F > 0 for the this interval.         The slope dU/dx between x2 and x3 is positive;  Thus F < 0 for the same interval 

The total mechanical energy is ( ) ( )This energy is constant (equal to 5 J in the figure)

Turning Points:

mecE K x U x= +This energy is constant (equal to 5 J in the figure)and is thus represented by a horizontal line. We can solve this equation for ( ) and gK x et:

( ) ( ) At any point on the -axiswe can read the value of ( ). Then we can solve th ti b d d t i

mecK x E U x x xU x

K

= −

the equation above and determine K2

From the definition of the kinetic energy cannot be negative. 2

mvK =

This property of K allows us to determine which regions of the x-axis motion is allowed. ( ) ( ) ( )If 0 ( ) 0

mecK x K x E U xK E U

= = −> → > ( ) M ti i ll dU E→ <If 0 ( ) 0mechK E U x> → − >

If 0 ( ) 0 The points at which: ( ) are known as

( )( )

turning point

Motion is aMotion is forbidden

llowed

smech

me

me

c

c

mec

U x EU xK E U x

E UE

x

→< → − < →

<

=>

1

p ( )for the motion. For example is the turning

g p point for ht

mec

x e versus plot above. At the turning point 0 U x K =

Given the U(x) versus x curve the turning points and the regions for which motion is allowed depends on the value of the mechanical energy Emec

In the picture to the left consider the situation when Emec = 4 J (purple line)  The turning points (E = U ) occur at x1 and x > x5. Motion is allowed for x > x1 If weturning points (Emec  U )  occur at x1 and x > x5.  Motion is allowed for   x > x1 If we reduce Emec to 3 J or 1 J the turning points and regions of allowed motion change accordingly.

Equilibrium Points: A position at which the slope dU/dx 0 and thus F 0 is calledEquilibrium Points: A position at which the slope dU/dx = 0 and thus F = 0 is called an equilibrium point. A region for which F = 0 such as the region x > x5 is called a region of neutral equilibrium.  If we set Emec = 4 J  the kinetic energy K = 0 and any particle moving under the influence of U will be stationary at any point with x > xparticle moving under the influence of U will be stationary at any point with    x > x5   

Minima in the U versus x curve are positions of stable equilibriumMaxima in the U versus x curve are positions of unstable equilibriumMaxima in the U versus x curve are positions of unstable equilibrium

Note: The blue arrows in the figure  indicate the direction of the force F as determined from the equation: 

( )( ) dU xF x =( )F xdx

= −

Positions of Stable Equilibrium. An example is point x4 where U has a minimum. If we arrange Emec = 1 J then K = 0 at point x4.  A particle  with          Emec = 1 J is stationary at x4.  If we displace slightly the particle either to the right or to the 4left of x4 the force tends to bring it back to the equilibrium position.  This equilibrium is stable.  

Positions of Unstable Equilibrium. An example is point x3 where U has aPositions of Unstable Equilibrium. An example is point x3 where U has a maximum. If we arrange Emec = 3 J then K = 0 at point x3.  A particle  with Emec = 3 J is stationary at x3.  If we displace slightly the particle either to the right or to the left of x3 the force tends to take it further away from the equilibrium 3 y qposition.  This equilibrium is unstable

Work done on a System by an External Force

( f i i i l d)(no friction involved)

Up to this point we have considered only isolated systems in which no external forces  were present.  We will now 

id t i hi h th f t l t thconsider a system in which there are forces external to the system

Th d d i b li b ll b i h l d b l Th i f hThe system under study is a bowling ball being hurled by a player.  The system consists of the ball and the earth taken together.  The force exerted on the ball by the player is an external force.  In this case the mechanical energy Emec of the system is not constant. Instead it changes by an amount equal to the work W done by the external force according to the equation: y q y g q

mecW E K U= Δ = Δ + Δ

Work done on a System by an External Force (friction involved)

A Bd

vo v•2nd Newton’s Law for x component F‐fk=ma

m m

A Bvofk fk

x

F•Since a=const, v2 = vo2 +2ad, solving for a and substituting the result in Newton’s Eq.

Fd=1/2mv2‐1/2mvo2+fkd=ΔK+fkd

Th h i fi d h h i i h l b lidi•Through experiment, we find that the increase in thermal energy by sliding : ΔEth=fkd 

•Fd is the work W done by external force F on the block floor systemy y

•Hence, W=ΔEmec+ΔEth energy statement for the work done on a system by an external force when friction is involved.

Problem 42. A worker pushed a 27 kg block 9.2 m along a level floor at constant speed with a force directed 32 degree below the horizontal. If the coefficient of kinetic friction between block and floor was 0.20, what were (a) the work done by the worker’s force and (b) the , ( ) y ( )increase in thermal energy of the block‐floor system?

. Since the velocity is constant, ra = 0 and the horizontal component of the worker'spush F cos θ (where θ = 32°) must equal the friction force magnitude fk = μk FN. Also, the vertical forces must cancel implyingvertical forces must cancel, implying FN.=Fsinθ +mg, on the other hand Fcos θ=μk FN which is solved to find F = 71 N. (a) The work done on the block by the worker is, using Eq. 7-7,

W Fd= = °= ×cos .θ 71 56 102 N 9.2 m cos32 J .b gb g (b) Since fk = μk (mg + F sin θ ), we find 2

th (60 N)(9.2m) 5.6 10 J.kE f dΔ = = = ×(b) S ce fk μk (mg s θ ), we d th ( )(9. ) . J.kf

Conservation of Energy

Energy cannot magically appear or disappear. It can be transferred to or from objects and systems by means of work done on a system. 

The total energy E of a system can change only by amounts of energy that are transferred to or from the system.

W=ΔE=ΔE +ΔE h+ΔEiW=ΔE=ΔEmec+ΔEth+ΔEint

The law of conservation of energy is not something we have derived from basic physics principles. Rather it is a law based on countless experiments.  We have never found an exception to itexception to it.

Isolated SystemIf a system is isolated from its environment, there can be no energy transfers to or from it The total energy E of an isolated system cannot changeit. The total energy E of an isolated system cannot change.

Problem 61. A stone with a weight of 5.29 N is launched vertically from ground level with an initial speed of 20.0 m/s, and the air drag on it is 0.265 N throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits theare (a) the maximum height reached by the stone and (b) its speed just before it hits the ground?

(a) The maximum height reached is h. The thermal energy generated by air resistance asthe stone rises to this height is ΔE h = fh by Eq 8 31 We use energy conservation in thethe stone rises to this height is ΔEth = fh by Eq. 8-31. We use energy conservation in theform of Eq. 8-33 (with W = 0):

K U E K Uf f i i+ + = +Δ th and we take the potential energy to be zero at the throwing point (ground level). The

initial kinetic energy is K mvi =12 0

2 , the initial potential energy is Ui = 0, the final kinetic

i K 0 d th fi l t ti l i U h h i th i ht fenergy is Kf = 0, and the final potential energy is Uf = wh, where w = mg is the weight of

the stone. Thus, wh fh mv+ =12 0

2 , and we solve for the height:

2 20 0mv vh = =

2( ) 2 (1 / )h

w f g f w= =

+ +.

Numerically, we have, with m = (5.29 N)/(9.80 m/s2)=0.54 kg,

2

2

(20.0 m/s) 19.4 m2(9.80 m/s )(1 0.265/5.29)

h = =+

.

(b) We notice that the force of the air is downward on the trip up and upward on the tripdown, since it is opposite to the direction of motion. Over the entire trip the increase in

1thermal energy is ΔEth = 2fh. The final kinetic energy is K mvf =12

2 , where v is the speed

of the stone just before it hits the ground. The final potential energy is Uf = 0. Thus, using Eq. 8-31 (with W = 0), we find

12

2 12

202mv fh mv+ = .

W b tit t th i f d f h t bt iWe substitute the expression found for h to obtain

22 20

02 1 1

2 (1 / ) 2 2fv mv mv

g f w= −

+

which leads to

2 22 2 2 2 20 0

0 0 0 02 2 21

(1 / ) (1 / )fv fv f w fv v v v v

mg f w w f w w f w f⎛ ⎞ −

= − = − = − =⎜ ⎟+ + + +⎝ ⎠

( ) ( )g f f f f⎝ ⎠

where w was substituted for mg and some algebraic manipulations were carried out.Therefore,

5 29 N 0 265 Nw f− −0

5.29 N 0.265 N(20.0 m/s) 19.0 m/s5.29 N 0.265 N

w fv vw f

− −= = =

+ +.

GENERAL PHYSICS PH 221-3A (Dr. S. Mirov) Test 2 (10/10/07)

STUDENT NAME: _Key____________________ STUDENT id #: ___________________________ -------------------------------------------------------------------------------------------------------------------------------------------

Sample Test 2

ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.

NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.)

Important Formulas:

1. Motion along a straight line with a constant acceleration vaver. speed = [dist. taken]/[time trav.]=S/t; vaver.vel. = Δx/Δt; vins =dx/Δt; aaver.= Δvaver. vel./Δt; a = dv/Δt; v = vo + at; x= 1/2(vo+v)t; x = vot + 1/2 at2; v2 = vo

2 + 2ax (if xo=0 at to=0)

2. Free fall motion (with positive direction ↑) g = 9.80 m/s2; y = vaver. t vaver.= (v+vo)/2; v = vo - gt; y = vo t - 1/2 g t2; v2 = vo

2 – 2gy (if yo=0 at to=0)

3 M ti i l3. Motion in a plane vx = vo cosθ; vy = vo sinθ; x = vox t+ 1/2 ax t2; y = voy t + 1/2 ay t2; vx = vox + at; vy = voy + at;

4. Projectile motion (with positive direction ↑) vx = vox = vo cosθ; x = v t;x vox t; xmax = (2 vo

2 sinθ cosθ)/g = (vo2 sin2θ)/g for yin = yfin;

vy = voy - gt = vo sinθ - gt; y = voy t - 1/2 gt2;

5. Uniform circular Motion a=v2/r, T=2πr/v

6. Relative motion P A P B B A

P A P B

v v va a

= +=

r r r

r r

7. Component method of vector addition

A = A1 + A2 ; Ax= Ax1 + Ax2 and Ay = Ay1 + Ay2; A A Ax y= +2 2 ; θ = tan-1 ⏐Ay /Ax⏐;

The scalar product A = c o sa b a b φ⋅rr

ˆ ˆˆ ˆ ˆ ˆ( ) ( )x y z x y za b a i a j a k b i b j b k⋅ = + + ⋅ + +

rr

r = x x y y z za b a b a b a b⋅ + +

r

The vector product ˆ ˆˆ ˆ ˆ ˆ( ) ( )x y z x y za b a i a j a k b i b j b k× = + + × + +rr

ˆˆ ˆˆˆ ˆy z x yx z

x y zy z x yx z

x y z

i j ka a a aa a

a b b a a a a i j kb b b bb b

b b b× = − × = = − + =

r rr r

ˆˆ ˆ( ) ( ) ( )

y

y z y z z x z x x y x ya b b a i a b b a j a b b a k= − + − + −

1. Second Newton’s Law ma=Fnet ;

2. Kinetic friction fk =μkN;

3 St ti f i ti f N3. Static friction fs =μsN;

4. Universal Law of Gravitation: F=GMm/r2; G=6.67x10-11 Nm2/kg2;

5. Drag coefficient 212

D C A vρ=

6. Terminal speed 2

tm gv

C Aρ=

7. Centripetal force: Fc=mv2/r

8. Speed of the satellite in a circular orbit: v2=GME/r

r9. The work done by a constant force acting on an object: c o sW F d F dφ= = ⋅

rr

10. Kinetic energy: 212

K m v=

11. Total mechanical energy: E=K+U

12. The work-energy theorem: W=Kf-Ko; Wnc=ΔK+ΔU=Ef-Eo

13. The principle of conservation of mechanical energy: when Wnc=0, Ef=Eo

14. Work done by the gravitational force: c o sgW m g d φ=

1. Work done in Lifting and Lowering the object:

; ; f i a g f i a gK K K W W i f K K W WΔ = − = + = = −

2. Spring Force: ( H o o k 's l a w )xF k x= −

3. Work done by a spring force: 2 2 21 1 1; i f 0 a n d ; 2 2 2s i o i f sW k x k x x x x W k x= − = = = −

4. Work done by a variable force: ( )fx

x

W F x d x= ∫ ix

5. Power: ; ; c o sa v gW d WP P P F v F v

t d tφ= = = = ⋅

Δ

r r

6. Potential energy: ; ( )f

i

x

xU W U F x d xΔ = − Δ = − ∫

i

7. Gravitational Potential Energy:

( ) ; 0 a n d 0 ; ( )f i i iU m g y y m g y i f y U U y m g yΔ = − = Δ = = =

8. Elastic potential Energy: 21( )U x k x=8. Elastic potential Energy: ( )2

U x k x

9. Potential energy curves: ( )( ) ; ( ) ( )m e c

d U xF x K x E U xd x

= − = −

10. Work done on a system by an external force:

F r i c t i o n i s n o t i n v o l v e d W h e n k i n e t i c f r i c t i o n f o r c e a c t s w i t h i n t h e s y s t e m

m e c

m e c t h

t h k

W E K UW E E

E f d

= Δ = Δ + Δ= Δ + Δ

Δ =

11 Conservation of energy: i n tm e c t hW E E E E= Δ = Δ + Δ + Δ11. Conservation of energy:

i n tf o r i s o l a t e d s y s t e m ( W = 0 ) 0m e c t hE E EΔ + Δ + Δ =

12. Power: ; a v gE d EP Pt d t

Δ= =

Δ

1.

4T4T

61.2 kg

16

22

‐2

=4x103 m4x10 m

6.

7. A 700 g block is released at height ho above a vertical spring of constant k=400 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 19.0 cm. How much work is done

(a) by the block on the spring? (b) by the spring on the block? (c) What is the value of ho? (d) If the block were released from height 2.00 ho above the spring, what would be the

i i f h i ?maximum compression of the spring?

We place the reference position for evaluating gravitational potential energy at therelaxed position of the spring. We use x for the spring's compression, measured positivelydownwards (so x > 0 means it is compressed). (a) With x = 0 190 m(a) With x = 0.190 m,

21 7 .2 2 J 7 .2 J2sW k x= − = − ≈ −

for the work done by the spring force Using Newton's third law we see that the workfor the work done by the spring force. Using Newton s third law, we see that the workdone on the spring is 7.2 J. (b) As noted above, Ws = –7.2 J. (c) Energy conservation leads to

K U K U

m g h m g x k x

i i f f+ = +

= − +021

2

which (with m = 0.70 kg) yields h0 = 0.86 m. (d) With a new value for the height ′ = =h h0 02 1 7 2. m , we solve for a new value of xusing the quadratic formula (taking its positive root so that x > 0).

m g h m g x k x xm g m g m g k h

k′ = − + ⇒ =

+ + ′0

22

012

2b g

which yields x = 0.26 m.

8. A particle moving along the x axis is acted upon by a single force F = F0e-kx, where F0 and k are constants. The particle is released from rest at x = 0. What is the maximum kinetic energy it will attain?

According to the work energy theorem since we started from rest ( 0)f i f iW K K K K= − = =fx

0 0 00

g gy ( )

1On the other hand ( ) ( ) = f

f f f

f i f i

kxkxx x x kxo o o

e eW F x dx K F x dx F e dx F Fk k k

−−− ⎛ ⎞⎡ ⎤

= → = = = − − +⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

∫ ∫ ∫' ' kMax occurs at a pointfK ' '

max

when ( ) 0, ( ) ( ) 0 only for x=kxo

o

K x K x F e

FK Joulesk

−= = = ∞

→ =

9. The potential energy of a 0.20-kg particle moving along the x axis is given by U(x) = (8.0J/m2)x2 + (2.0J/m4)x4 . When the particle is at x = 1.0m it is traveling in the positive x direction with a speed of 5.0m/s. At what x coordinate it will stop momentarily to turn around.p p y

2 2

( ) Calculate total mechanical energy of the particle at coordinate x=1.0 m0.2 5.0(1.0 ) 8.0 2.0 10.0 ; (1.0 ) 2.5

2 2

amvU m J J J K m J×

= + = = = =

12.5( ) The particle will stop momentarily to turn around when (

mecE Jb K

→ =

2 4

) ( ) 0

8.0 2.0 12.5mecx E U x

x x

= − =

→ + =2

2

2 2

8.0 2.0 12.5

8.0 8.0 4(2.0)(12.5)4.0

( 2 0 3 2)

x x

x

x m

→ +

− ± +=

= − ±2 2

( 2.0 3.2)1.2

1.1

x mx mx m

= − ±

==