GENERAL PHYSICS PH 221-3A (Dr. S. Mirov) Test 2 (02/27/13)

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ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FOUR 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. Integration in Motion Analysis (non-constant acceleration) 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; 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;

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

6. Uniform circular Motion a=v2/r,

1. Relative motion

P A P B B A

P A P B

v v va a

2. Component method of vector addition

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

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/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

5. Work done by a variable force: ( )f

i

x

x

W F x d x

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

t d t

7. Potential energy: ; ( )f

i

x

xU 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

9. Elastic potential Energy: 21( )2

U 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:

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

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.

5

2. A block of mass m=5 kg starting from rest slides (from the top) down an incline planeof length 15 m and angle =37. Coefficient of friction is k=0.30.

a) Show free body diagram and calculate acceleration of the block. b) Calculate the velocity of the block when it reaches the bottom of the incline. c) Calculate work done by the friction force when the block reaches the bottom of the

incline. d) Calculate the velocity of the block when it reaches the bottom of the incline using

work-energy theorem.

x

Ny

fk

mg

mgsin

mgcos

2

2 1

(a) y: cos 0 x: sin (sin cos ) cos

(sin cos ) (9.8)(sin 37 0.30cos37) 3.55 4

( ) 0 2 2 2(3.55)(15) 10.3 1 10

( ) ( cos ) (0.k

k

k

f k

N mgmg f ma mg maf N mg

ma gs

mb v ax v axs

c W f x mg x

2

2 2

1

3)(5)(9.8)(cos37)(15) 176.1 2 10

( ) 0 0; sin ;2 2

2 ( 176.1)(2)2 sin 2(9.8)(15)sin 37 10.3 1 105

k k k

k

f f f i i f f

f

J

mv mvd W U K U K mgh mgh W mgx W

W mv gxm s

3. In Fig., (a) How large is a force F is needed to give the blocks the acceleration of

3.0 m/s2 if the coefficient of friction between blocks and the table is0.20?

(b) How large a force does the 1.5 kg block then exert on the 2.0 kg block?

F

1.5 kg 1.0 kg 2.0 kg

F

N

Mg

fk

Y +

X +

1 2 3

1 2 3a) x: ; where ; ( ) 4.5(3 0.2 9.8) 22 y: 0;

k

k

F f Ma M m m m F Ma Mg M a g NN Mg N Mgf N

F

N1

m1g

fk1

Y +

X +

F21

1 21 1 21 1 1 1 1 1

1 1 1

1 1

12 21

b) x: ; ( ) 22 1.5(0.2 9.8 3) 15 y: 0; According to the third Newton's Law 15

k k

k

F f F m a F F f m a F m g m a F m g aN

N m g N m gf N

F F N

4.

Since the valley is frictionless, the only reason for the speed being less when it reachesthe 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 where 2 / 2i iK mv and vi = 6.0 m/s. Dividing by mass and rearranging, we obtain

d vg

hi 2

212

. m.

6.

7.