P6.5 (a) static friction. v r. = r ( 30.0 cm )( 980 cm s ) P6.15 Let the tension at the lowest point be T.

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1 Q65 The speed chanes The tanential orce component causes tanential acceleration Q69 I would not accept that statement or two reasons First, to be beyond the pull o raity, one would hae to be ininitely ar away rom all other matter Second, astronauts in orbit are moin in a circular path It is the raitational pull o Earth on the astronauts that keeps them in orbit In the space shuttle, just aboe the atmosphere, raity is only slihtly weaker than at the Earth s surace Graity does its job on an orbitin spacecrat, because the crat eels no other orces and is in ree all P65 (a) static riction (b) mai ˆ = i ˆ + n ˆ j+ m( ˆ j ) Fy = = n m thus n= m and Fr = m = = µ n= µ m r ( 5 cm s) Then µ = = = r ( cm )( 98 cm s ) 85 P65 Let the tension at the lowest point be T m F= ma: T m= mac = r T = m + r ( 8 m s) T = ( 85 k) 98 m s + = 8 kn > N m P65 (a) He doesn tm ake itacross the rier because the ine breaks m Fy = may = (b) When n =, m m n= m m= Then, = m n= m FIG P65 P658 (a) Since the object o mass m is in equilibrium, F y = T m = or T = m (b) The tension in the strin proides the required centripetal acceleration o the puck (c) Thus, Fc = T = m From P6 5 N horizontally inward P6 6 N P66 (a) 65 km s; (b) 68 s P ˆ + 6 ˆm s P6 6 re m in P6 (a) T ; (b) T upward m P66 (a) m s ; (b) P N P6 (a) 86 m; (b) downward; (c) m Fc = we hae P6 (a) ( i j ) ; (b) 65 m s ; (c) ( 8 i+ 8 j ) m s 79 m s orward and 8 inward ˆ F m = = m ˆ c m 85 m s, Unless they are belted in, the riders will all rom the cars

2 P6 5 m s Straiht across the dashboard to the let P6 57 P66 (a) h; (b) 7 P68 ( t L) µ k = ( + a) t P6 5 (a) 8 m s horizontally inward = ; (b) 6 N inward perp to cone; (c) 75 m s P6 (a) 67 m s downward; (b) 78 N up; (c) 8 N up P6 (a) 58 m s; (b) 8 m P66 P68 m s P6 65 m s P6 (a) 98 m s; (b) see the solution P66 (a) 77 k m ; (b) 998 N; (c) The ball reaches max heiht 9 m Its liht lasts 6 s Impact speed 7 m s P68 (a) see the solution; (b) 88 m; (c) 59 P65 85 re s P65 m (a) m ; (b) = P65 (a) 6 m s ; (b) m; (c) 77 m s P656 (a) 6 N; (b) 96 P658 (a) m ; (b) m ; (c) m m P66 6 re m in P66 re m in P66 (a) = π; (b) mπ P666 (a) 8 s; (b) 79 m s; (c) 9 cm s; (d) 955 cm P668 (a) either 7 or ; (b) P67 (a) 78 m s; (b) s; (c) m Q76 No The ectors miht be in the third and ourth quadrants, but i the anle between them is less than 9 their dot product is positie Q79 The rock increases in speed The arther it has allen, the more orce it miht exert on the sand at the bottom; but it miht instead make a deeper crater with an equal-size aerae orce The arther it alls, the more work it will do in stoppin Its kinetic enery is increasin due to the work that the raitational orce does on it P77 (a) W F r F x F y = = + = 6 N m + N m = 6 J x y *P7 (b) F r 6 = cos = cos = 69 F r (( 6) + ( ) ) ( ) + ( ) The same orce makes both liht sprins stretch (a) The hanin mass moes down by m m x= x+ x = + = m k k + k k m m = 5 k 98 m s + = m N 8 N (b) We deine the eectie sprin constant as

3 *P7 See the solution to problem 7 (a) x m = + k k (b) k = + k k P7 (a) W kx kx F m k = = = + x m k + k k k ( ) m m = + = N 8 N s= i = 5 5 = 65 J W s= m mi = m so ( W ) 65 = = m s = 79 m s m 7 N m (b) mi k x+ W s= m ( 5)( )( 98)( 5 ) J+ 65 J= m 8 J= ( k) P78 (a) 8 = m s = 5 m s i ˆ= cosα But also, i ˆ= x FIG P7 P76 Thus, ( ) and cosα = x or cosα = x Similarly, cos β = cos γ = z where x y z = + + y z x (b) cos α + cos β+ cos γ = + + = = K + W + W = K i s mi + kxi kx+ m xcos = m FIG P76 + kxi + m x icos = m ( N cm )( 5 cm )( 5 m ) ( k)( 98 m s ) ( 5 m ) sin = ( k) 5 J 85 J= 5 k = = 5 68 m s P7 59 J P7 (a) 8 J; (b) 8 J P78 5 W P7 6 P7 (a) see the solution; (b) J P7 5 J P76 (a) 575 N m ; (b) 6 J P78 (a) 9 kj; (b) 7 kj, larer by 96% P7 (a) see the solution; (b) m y

4 P7 m m (a) + ; (b) k k + k k P7 (a) J; (b) 5 m s; (c) 6 J P76 (a) 6 J; (b) 6 J P78 (a) 9 m s; (b) 5 m s; (c) 87 m s P7 6 (a) 78 J; (b) 5 N ; + 6 (c) 8 m s ; (d) 9 ns P7 (a) 79 m s; (b) 5 m s P7 (a) 9 J; (b) ; (c) ; (d) 85 J; (e) J P76 8 W P78 ~ W P7 (a) 59 kw; (b) kw P7 No (a) 58; (b) 95 W = hp Q8 The inal speed o the children will not depend on the slide lenth or the presence o bumps i there is no riction I there is riction, a loner slide will result in a lower inal speed Bumps will hae the same eect as they eectiely lenthen the distance oer which riction can do work, to decrease the total mechanical enery o the children Q85 Kinetic enery is reatest at the startin point Graitational enery is a maximum at the top o the liht o the ball P85 U i+ Ki= U + K : mh+ = m + m ( 5) = + = F= m : n+ m= m *P86 n= m = m m = n = 5 k 98 m s = 98 N downward FIG P85 useuloutputenery useuloutputpow er eiciency = = totalinputenery totalinputpow er ( ) air ( l ) mwatery t ρ t y ρ t y e= = = m t r t water w ater w w ρair π ρπ a r where l is the lenth o a cylinder o air passin throuh the mill and w is the olume o water pumped in time t We need inject neliible kinetic enery into the water because it starts and ends at rest w eρπ 75 k m 5 m m s a r π = = t ρ y k m 98 m s 5 m P8 U i+ Ki+ Emech = U + K : w L 6 s = 66 m s = 6 L m in m min mh h= m + m = µ n= µ m mh µ mh= m+ m ( µ ) m m h = m + m FIG P8

5 98 m s 5 m 5 k k = = 8 k P85 (a) F x is zero at points, C and E; F x is positie at point B and neatie at point D (b) and E are unstable, and C is stable (c) F x 7 m s B C E x (m) D P88 The potential enery o the block-earth system is mh n amount o enery µ kmdcos is conerted into internal enery due to riction on the incline Thereore the inal heiht y max is ound rom mymax = mh µ kmdcos where ymax d = sin mymax = mh µ kmymaxcot y max Solin, h h ymax = + µ k cot FIG P88 P86 Let λ represent the mass o each one meter o the chain and T represent the tension in the chain at the table ede We imaine the ede to act like a rictionless and massless pulley (a) For the ie meters on the table with motion impendin, F y = : + n 5 = n= 5λ s µ sn= 65 ( λ) = λ F x = : + T s = T = s T λ The maximum alue is barely enouh to support the hanin sement accordin to F y = : + T = T = λ so it is at this point that the chain starts to slide (b) Let x represent the ariable distance the chain has slipped since the start Then lenth ( 5 x) remains on the table, with now F y = : + n ( 5 x) λ= n= ( 5 x) λ = µ n= 5 ( x) λ= λ xλ k k Consider eneries o the chain-earth system at the initial moment when the chain starts to slip, and a inal moment when x = 5, when the last link oes oer the brink easure heihts aboe the inal position o the leadin end o the chain t the moment the inal link slips o, the center o the chain is at y = meters Oriinally, 5 meters o chain is at heiht 8 m and the middle o the danlin sement is at heiht 8 = 65 m + + = + : ( ) k K U E K U i i mech + m y + m y dx = m + m y i 5 i

6 P8 (a) 8 J; (b) 7 J; (c) P8 9 (a) J; (b) P86 8 m P88 (a) kw; (b) 6 kw; (c) P8 kx d= x msin P8 (a) see the solution; (b) 6 P8 (a) ( m m) h mh ; (b) ( m+ m ) m + m P86 6 L m in P J ( λ) + ( λ) ( λ xλ) dx= ( λ) + ( λ) 6 ( 5 m )( 98 m s ) dx+ xdx= x 75 x + = = 5= = = P8 8h 5 P8 (a) see the solution; (b) 5 J P8 (a) B = 59 m s; C = 767 m s; (b) 7 J P86 (a) U = J; E = J; (b) Yes The total mechanical enery chanes P88 9 m P8 6 kn up P8 68 J P8 (a) 5 m s; (b) yes; (c) 6 m; (d) ir dra depends stronly on speed P86 9 kj P88 kw x Bx 5 9B 9B 5 P8 (a) ; (b) U = ; K = P8 ( 7 9xyˆ ˆ ) i x j P86 (a) r= 5 mm and mm, stable; mm and unstable; r neutral; (b) 56 J E < J; (c) 6 mm r 6 mm ; (d) 6 J; (e) 5 mm; () J P85 kw P85 (a) 588 J; (b) 588 J; (c) m s; (d) 96 J; 9 J P85 5 P856 (a) J; (b) m; (c) 8 m s; (d) 98 mm ; (e) 85 m s ˆ P858 (a) ( x x ) i; (b) 87; -55; P86 (a) 78 m; (b) m s; (c) 8 m P86 (a) see the solution; (b) 7 m s P86 (a) see the solution; (b) 5 m s; (c) 958 m s; (d) see the solution P866 9 m s P868 m 7 m s

7 P87 6 P87 (a) m s; (b) 79 J; (c) 8 N; (d) 77 N; (e) 57 kn up Q9 omentum conseration is not iolated i we choose as our system the planet alon with you When you receie an impulse orward, the Earth receies the same size impulse backwards The resultin acceleration o the Earth due to this impulse is siniicantly smaller than your acceleration orward, but the planet s backward momentum is equal in manitude to your orward momentum Q9 The planet is in motion around the sun, and thus has momentum and kinetic enery o its own The spacecrat is directed to cross the planet s orbit behind it, so that the planet s raity has a component pullin orward on the spacecrat Since this is an elastic collision, and the elocity o the planet remains nearly unchaned, the probe must both increase speed and chane direction or both momentum and kinetic enery to be consered P99 p= F t *P9 p = m = m cos6 mcos6 = y y iy px = m sin6 sin6 = msin6 = ( k )( m s)( 866 ) = 5 k m s px 5 k m s Fae = = = 6 N FIG P99 t s For the car-truck-drier-drier system, momentum is consered: pi+ pi= p+ p : ( k) ( 8 m s) ˆ i+ ( 8 k) ( 8 m s)( ˆ i) = ( 8 k) ˆ i 5 6 k m s = = 5 m s 8 k For the drier o the truck, the impulse-momentum theorem is t= F s = 8 k 5 m s i ˆ 8 k 8 m s i ˆ F p p i: ˆ F= 78 N ( i ) on the truck drier For the drier o the car, F( s) = ( 8 k)( 5 m s) i ˆ ( 8 k)( 8 m s ˆ )( i ) ˆ F= 889 N i on the car drier, 5 times larer P9 By conseration o momentum or the system o the two billiard balls (with all masses equal), 5 m s+ = ( m s) cos + x = 5 m s x y = m s sin + = 6 m s y = 5 m s at 6 FIG P9 Note that we did not need to use the act that the collision is perectly elastic P9 Let represent the area o the bottom row o squares, the middle square, and the top pair 7

8 = cm, = + + = + + = = cm, cm = = = 6 cm cm = = = 6 cm 6 cm = = = 6 cm x x y y C C C C = cm, = 6 cm ( 5 cm ) + ( 5 cm ) + ( 5 cm ) 6 ( ) + ( ) + ( ) x + x 5 cm 5 cm cm + x = = = 7 cm 6 = = cm = cm P958 Usin conseration o momentum rom just beore to just ater the impact o the bullet with the block: + m mi= ( + m) or i= m The speed o the block and embedded bullet just ater impact may be ound usin kinematic equations: d= tand h= t h Thus, t= and d d = = d = t h h i m h FIG P9 d Substitutin into () rom aboe ies i + m = m P967 (a) Find the speed when the bullet emeres rom the block by usin momentum conseration: d h m/s m = V + m i i The block moes a distance o 5 cm ssume or an approximation that the block quickly reaches its maximum elocity, V i, and the bullet kept oin with a constant elocity, The block then compresses the sprin and stops 5 cm V i = kx ( 9 N m )( 5 m ) Vi = = 5 m s k mi Vi = = m = m s 8 ( 5 k)( m s) ( k)( 5 m s) 5 k

9 ( 9 N m )( 5 + m ) E = 7 J, or there is an enery loss o 7 J (b) E= K + U = ( 5 k)( m s) ( 5 k)( m s) P9 (a) ; (b) 6 k m s ; upward P9 (a) 6 m s to the let; (b) 8 J P96 The orce is 6 kn P98 9 k m s upw ard P9 (a) 5 N s toward the net; (b) 7 J P9 ~ N upward P9 (a) and (c) see the solution; (b) small; (d) lare; (e) no dierence P96 67 m s P98 (a) 5 m s; (b) 75 J P9 556 m P9 78 kn on the truck drier; 889 kn in the opposite direction on the car drier P9 = m l P96 79 cm P98 (a) 88 m s at ; (b) 78 J becomes internal enery P9 Y = i sin ; O = icos P9 No; his speed was 5 m i h P9 (a) = i ; (b) 5 and 5 P96 (a) i ; P98 (, m ) 6 i ; (b) 5 P9 67 m rom the Earth s center P9 8 (a) see the solution; (b) 57 J P9 6 5L P96 (a) see the solution; (b) ( m, m ) ; (c) ( ˆi ˆj ) m s; (d) ˆ ( i ˆj ) 5 5 k m s P98 (a) 78 ˆi m s; ˆi m s; (b) 6 ˆi m s P95 (a) 787 m s; (b) 8 m s P95 m+ m m (a) ;(b) m+ m m ;(c) km ( + m ) m m m = + m+ m m+ m m m m = + m+ m m+ m P956 9 N P958 + m d m h P96 (a) 667 m s; (b) 95 m P96 (a) 68 m s; (b) m P96 (a) 5 m s; (b) 77 m; (c) 5 kn; (d) No The rails exert a ertical orce to chane the momentum P966 N to the riht P m s P97 (a) 7 km s; (b) 5 km P97 (a) 75 N to the riht; (b) 75 N to the riht; (c) 75 N; (d) 8 J; (e) J; () Friction between sand and belt conerts hal o the input work into extra internal enery 9