GENERAL PHYSICS PH 221-3A (Dr. S. Mirov) Test 2 (10/06/10)

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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 acceleration1. Motion along a straight line with a constant accelerationvaver. 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. Integration in Motion Analysis (non-constant acceleration) 1t 1

1

o

t

ot

v v a d t

1

1

o

t

ot

x x v d t

4. Motion in a plane vx = vo cos; x o ;vy = vo sin; x = vox t+ 1/2 ax t2; y = voy t + 1/2 ay t2; vx = vox + at; vy = voy + at;

5. 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;

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

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

1. Relative motion

P A P B B A

P A P B

v v va a

2 Component method of vector addition2. Component method of vector additionA = 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

Th t d t ˆ ˆˆ ˆ ˆ ˆ( ) ( )a b a i a j a k b i b j b k

The vector product ( ) ( )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

3. Second Newton’s Law: ma=Fnet ;

4. Kinetic friction fk =kN;

5. Static friction fs =sN;

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

7. Drag coefficient 212

D C A v

8. Terminal speed 2

tm gv

C A

9. Centripetal force: Fc=mv2/r9. Centripetal force: Fc mv /r

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

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

12. Kinetic energy: 212

K m v

13. Total mechanical energy: E=K+U

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

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

1. Work done by the gravitational force: c o sgW m g d

2. 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

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

4. 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

( )fx

W F d5. Work done by a variable force: ( )ix

W F x d x

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

t d t

7 P t ti l ( )fxU W U F d 7. Potential energy: ; ( )

ixU W U F x d x

8. 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

19. Elastic potential Energy: 21( )

2U x k x

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

d U xF x K x E U xd x

11. Work done on a system by an external force: y y

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

i tt hW E E E E 12. 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

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

1.

2.

F‐F21‐f1k=m1aF F f m a

=22 N

f1k

F F21 f1k m1aF‐F21‐f1k=m1aF21=F‐f1k‐m1a=22.3‐0.2x1.5x9.8‐1.5x3=15N

15 N15 N15 N

6. In figure below, a block slides along a track from one level to a higher level after passingthrough an intermediate valley. The track is frictionless until the block reaches the higherlevel. There a frictional force stops the block in a distance d. The block’s initial speed is

i 6 0 / th h i ht diff h i 1 1 d i 0 60 Fi d d?vo is 6.0 m/s, the height difference h is 1.1 m, and µk is 0.60. Find d?

Since the valley is frictionless, the only reason for the speed being less when it reachesy y p gthe higher level is the gain in potential energy U = mgh where h = 1.1 m. Sliding along the rough surface of the higher level, the block finally stops since its remaining kineticenergy has turned to thermal energyE f d mgdkth , where 0.60 . Thus, Eq. 8-33 (with W = 0) provides us with an equation to solve for the distance d:

K U E mg h di th b g

2where 2 / 2i iK mv and vi = 6.0 m/s. Dividing by mass and rearranging, we obtain

d v hi 2

12 mdg2

12

. m.