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

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

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. = x/t; vins =dx/t; aaver.= vaver. vel./t; a = dv/t; 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. Motion in a plane vx = vo cos; vy = vo sin; 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 cos; x = vox t; xmax = (2 vo

2 sin cos)/g = (vo2 sin2)/g for yin = yfin;

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

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

6. Relative motion P A P B B A

P A P B

v v va a

7. Component method of vector addition

Sample Test 2

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

The scalar product A = c o sa b a b

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

= x x y y z za b a b a b a b

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

ˆˆ ˆˆˆ ˆ

ˆˆ ˆ( ) ( ) ( )

y z x yx zx y z

y z x yx zx y z

y z y z z x z x x y x y

i j ka a a aa a

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

b b b

a 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 =kN;

3. Static friction fs =sN;

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

5. Drag coefficient 212

D C A v

6. Terminal speed 2

tm gv

C A

7. Centripetal force: Fc=mv2/r

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

9. The work done by a constant force acting on an object: c o sW F d F d

10. Kinetic energy: 212

K m v

11. Total mechanical energy: E=K+U

12. The work-energy theorem: W=Kf-Ko; Wnc=K+U=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

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 W

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: ( )f

i

x

x

W F x d x

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

t d t

6. Potential energy: ; ( )f

i

x

xU W U F x d x

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 y

8. Elastic potential Energy: 21( )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

11. Conservation of energy: i n t

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 h

m e c t h

W E E E EE E E

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

61.2 kg

4T4T

1.

5

2

-2

=4x103 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

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,

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 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?

fx

0 0 00

According to the work energy theorem since we started from rest ( 0)

1On the other hand ( ) ( ) =

Max occurs at a point

ff f f

f i f i

kxkxx x x kxo o o

f

W K K K K

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

K

' '

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.

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 12.5

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

amvU m J J J K m J

E Jb K

2 4

22

2 2

2 2

) ( ) 0

8.0 2.0 12.5

8.0 8.0 4(2.0)(12.5)4.0

( 2.0 3.2)1.2

1.1

mecx E U x

x x

x

x mx mx m