Homework: 5, 9, 19, 25, 31, 34, 39 (p )

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1 Hoework: 5, 9, 19, 5, 31, 34, 39 (p )

2 5. A 3.0 kg block is initilly t rest on horizontl surfce. A force of gnitude 6.0 nd erticl force P re then pplied to the block. The coefficients of friction for the block nd surfce re s =0.40 nd k =0.5. Deterine the gnitude of the frictionl force cting on the block if the gnitude of P is () 8.0, (b) 10, nd (c) 15 (1, textbook). g g k s, x s () P=8.0, =1.4 : s, x ks g P () friction 8.56 () s, x the block issttionry, therefore : friction 6.0 () s, x () (b) P=10.0, =19.4 : s,x friction (c) P=15, =14.4 : s,x < the block oes: friction k the block does not oe, 6.0 () s, x μ k g therefore : 5.76 () 3.6 ()

3 9. A 3.5 kg block is pushed long horizontl floor by force of gnitude 15 t n ngle =40 0 with the horizontl force. The coefficient of kinetic friction between the block nd the floor is 0.5. clculte the gnitude of () the frictionl force on the block fro the floor nd (b) the block s ccelertion. k k g k k sinθ 43.9 () () k g y x cosθ - k cosθ - k 0.14 (/s )

4 19. A 1 horizontl force pushes block weighing 5.0 ginst erticl wll. The coefficient of sttic friction between the wll nd the block is 0.6, nd the coefficient of kinetic friction is 0.4. Assue tht the block is not oing initilly. () Will the block oe? (b) In unit-ector nottion, wht is the force on the block fro the wll? () (b) s, x ks 0.61 g s,x wll wll g 5 () the block does ( s ( 1 0)î,x (5 7. () not oe, 0)ĵ s,x )î s ( g,y ( 1)î g friction 5 () s, y )ĵ (5)ĵ

5 5. Block B in the figure below weighs 750. The coefficient of sttic friction between block nd tble is 0.5; ngle is 30 0 ; ssue tht the cord between B nd the knot is horizontl. ind the xiu weight of block A for which the syste will be sttionry. y Here, we need to find the xiu lue of g,a. If the syste is sttionry: or block A: or block B: or Knot K: T g, A TA B T T B A ; s g,b ; Tcosθ Tsinθ s,x g,a is xi when s = s,x : T B k s T B s, x s Knot T A g,b T B T T A g,a () x g, A TA TBtn( ) tn(30 ) ()

6 31. Two blocks of weights 3.6 nd 7., re connected by ssless string nd slide down 30 0 inclined plne. The coefficient of kinetic friction between the lighter block nd the plne is 0.10; tht between the heier block nd the plne is 0.0. Assuing tht the lighter block leds, find () the gnitude of the ccelertion of the blocks nd (b) the tension in the string. or block A: A or block B: sin θ T f g, A A f k cosθ sin θ T f g, B B f k cosθ (1) B & () B A g, B 3.5 g, A ( (/s g,a ); B A g,b () A (1) )sinθ T 0.,A x B () y (f T g,a A f A f T,B B ) g,b f B

7 34. A slb of ss 1 = 40 kg rests on frictionless floor, nd block of ss =1 kg rests on top of the slb. Between block nd slb, the coefficient of sttic friction is 0.60, nd the coefficient of kinetic friction is The block is pulled by horizontl force of gnitude 10. In unit-ector nottion, wht re the resulting ccelertions of () the block nd (b) the slb? or the slb, long Ox: or the block, Ox: f f 1 1 f s, x s s g () f s, x therefore the block does slide on the slb μ - μ g k g k 1 1. (/s x 6.1 (/s ); 1 y ); (1. /s f (6.1 /s )î f ) î

8 39. Clculte the rtio of the drg force on jet flying t 100 k/h t n ltitude of 15 k to the drg force on prop-drien trnsport flying t hlf tht speed nd ltitude. The density of ir is 0.38 kg/ 3 t 15 k nd 0.67 kg/ 3 t 7.5 k. Assue tht the irplnes he the se effectie cross-sectionl re nd drg coefficient C. D 1 C ρ A R D D jet propeller ρ 15k 7.5k jet prop

9 Chpter orce nd Motion.1. ewton s irst Lw nd Inertil res.. ewton s Second Lw.3. Soe Prticulr orces. The Grittionl orce nd Weight.4. ewton s Third Lw.5. riction nd Properties of riction. Motion in the Presence of Resistie orces.6. Unifor Circulr Motion nd on-unifor Circulr Motion.5. Motion in Accelerted res

10 .6. Unifor Circulr Motion nd on-unifor Circulr Motion Unifor circulr otion Centripetl (rdil) ccelertion: R Centripetl (rdil) force: R ote: A centripetl force ccelertes n object by chnging its elocity direction without chnging its speed. T R

11 on-unifor circulr otion r t Rdil (centripetl) ccelertion The pth of prticle s otion r Tngentil ccelertion t r r ( t r r ) t ; t t r t r R ; t d dt

12 Sple Proble (p. 15) Diolo is riding bike in loop, ssuing the loop is circle with R =.7, wht is the lest speed Diolo cn he t the top of the loop to rein in contct with it there? g ( ) R g R To rein in contct with the loop: 0 the lest speed needed for the Diolo nd his bike: 0 in gr in ( / s) A free-body digr

13 Sple Proble (p. 18) Cured portions of highwys re tilted to preent crs fro sliding off the highwys. If the highwys re wet, the frictionl force fro the trck is negligible. Wht bnk ngle preents sliding? To preent sliding, the coponent r of the norl force long the rdil xis r proides the necessry centripetl force nd rdil ccelertion: r tn - sin cos 1 gr g - R to preent sliding Cr on leel trck Cr on bnked frictionless trck

14 .5. Motion in Accelerted res Accelerted (noninertil) reference fres: in which ewton s lws of otion do not hold. Exple: An eletor cb is oing with n 0 ccelertion the cb is not n inertil fre. + We choose the ground to be our inertil fre (sttionry), so using ewton s second lw for the pssenger with ss : g 0 + Howeer, if we choose the cb (noninertil fre, ccelerted with g 0 ) to be our fre, the pssenger s ccelertion is zero in this 0 fre, so In this cse, to use ewton s second lw, we ust dd n inertil (fictitious) force: fictitious 0

15 g 0 If the pssenger oes with n ccelertion g 0 In noninertil fre, ewton s second lw is: 0 0 in the cb:

16 Sple Proble (p. 103): In the figure below, pssenger of ss =7. kg stnds on pltfor scle in n eletor cb. We re concerned with the scle redings when the cb is sttionry nd when it is oing up or down. () ind generl solution for the scle reding, whteer the erticl otion of the cb. The scle reding is equl to the gnitude of the norl force cting on the pssenger. g (g )

17 (b) Wht does the scle red if the cb is sttionry or oing upwrd t constnt 0.50 /s? 0 (g ) () (c) Wht does the scle red if the cb ccelertes upwrd t 3.0 /s nd downwrd t 3.0 /s? (g (g (d) During the upwrd ccelertion in prt (c), wht is the gnitude net of the net force on the pssenger, nd wht is the gnitude p,cb of his ccelertion s esured in the fre of the cb? Does? net ) ) net p,cb 7.(9.8 3.) 7.(9.8 3.) g () p, cb 0 The pssenger is sttionry in the eletor, so: net p,cb 939 () 477 () The cb is not n inertil fre, hence ewton s second lw is not pplicble in the fre of the cb:

18 If we wnt to use ewton s second lw, we need to include fictitious force: net fictitious net cb,ground p,cp Hoework: 49, 51, 70 (p ) = 0 = 0

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