Conservation of Energy

Transcription

1 Chapter 8 Conseraton o Ener 8.3 U + K = U + K mh + = m ( ) + m ( 3.5 ) = ( ) + F= m = 3. n+ m= m 3. n = m = m =.m 3 n = m s =.98 N downward FIG. 8.3 (5. 3.) Δ A B 8.4 (a) K = W = W = m Δ h = m mb m A = m( 9.8)(.8) B = 5.94 m s A 5. m 3. m B C. m Smlarl, A = = 7.67 m s C FIG. 8.4 W A = m 3. m = 47 J C K + U = K + U 8. (a) m + = m + m mx + m = mx + m Note that we hae used the thaorean theorem to express the ornal netc ener n terms o the eloct components. Knetc ener tsel does not hae components. x = x Now, so or the rst ball ( sn 37. ) 4 = = =.85 m ( 9.8) and or the second ( ) 4 = = 5. m 9.8

2 The total ener o each s constant wth alue (. )( m s ). 7 = J 8. (a) For a 5-m cord the sprn constant s descrbed b F = x, m = (.5 m ). For a loner cord o lenth L the stretch dstance s loner so the sprn constant s smaller n nerse proporton 5 m m = = 3.33 m L L.5 m ( K+ U + Us) = ( K+ U + Us) + m + = + m + x m m( ) = x = 3.33 x L = 55 m = L+ x here 55. ml = 3.33 ( 55. m L) ml = 5.4 m 83 ml+.67l 3 =.67L 38 L+ 5.4 = ± ± L = = = m onl the alue o L less than 55 m s phscal. m = m xmax = x = 55. m 5.8 m = 9. m F= ma + x max m = ma m m m = ma 5.8 m a=.77= 7. m s ntal nal FIG. 8.(a) 8.3 F = ma n 39 N = n = 39 N = μ n= (.3)( 39 N ) = 8 N (a) (c) W = FΔ rcosθ = 3 5. cos = 65 J F Δ Ent = Δ x= 8 5. = 588 J W = nδ rcosθ = cos9 = n FIG. 8.3

3 (d) (e) W = mδ rcosθ = cos 9 = Δ K = K K = W ΔE other m = 65 J 588 J+ + = 6. J nt () K 6. J = = = m m s 8.8 (a) U = =. J U = K K + U E = 4. J Yes, Δ Emech = Δ K+ΔU s not equal to zero, some nonconserate orce or orces must act. For conserate orces Δ K+Δ U =. 8.9 U + K +Δ Emech = U + K mh h= m + m = μn= μm mh μmh = ( m+ m) ( μ ) m m h = m + m FIG m s.5 m = = m s 8. (a) Δ K = m( ) = m = 6 J (c) Δ U = m 3. m sn 3. = 73.5 J The mechancal ener conerted due to rcton s 86.5 J 86.5 J = = 8.8 N 3. m FIG. 8. (d) = μ n= μ mcos3. = 8.8 N 8.8 N μ = = m s cos3..679

4 8.5 (a) The object moed down dstance. m + x. Choose K + U + U +Δ E = K + U + U s mech s + m + + = + + x = 6 N m 4.7 N x 7.6 J (.5 )( 9.8 m s )(. m + x) = ( 3 N m ) x 6 ( Nm) 4.7 N ± 4.7 N 4 6 N m 7.6 N m x = 4.7 N ± 7 N x = 3 N m = at ts lower pont. The neate root tells how hh the object wll rebound t s nstantl lued to the sprn. We want x =.38 m x From the same equaton, (.5 )(.63 m s )(. m + x) = ( 3 N m ) = x x x The poste root s x =.43 m. (c) The equaton expressn the ener erson o the nonsolated sstem model has one more term m Δ x = x (.5 )( 9.8 m s )(. m + x).7 N (. m + x) = ( 3 N m ) x 7.6 J+ 4.7 N x.84 J.7 N x = 6 N m x x x = ± x = 3 x =.37 m 8.36 (a) Burnn lb o at releases ener cal 4 86 J = lb cal 7 lb.7 J The mechancal ener output s 7 (.7 J)(.) = nfδr cosθ

5 Then J nm cos = Δ J= n m s 8 steps.5 m J= n 5.88 J J n = = 3 where the number o tmes she must clmb the steps s 5.88 J Ths method s mpractcal compared to lmtn ood ntae. 58. Her mechancal power output s 3 W 5.88 J hp = = = 9.5 W = 9.5 W =. hp t 65 s 746 W ( K+ U) = ( K+ U) 8.37 (a) A B + ma = m B+ B = A = ( 9.8 m s ) 6.3 m =. m s (. m s) ac = = = r 6.3 m 9.6 m s up (c) F = ma + nb m= mac B m s 9.6 m s.3 N up n = + = (d) We compute the amount o chemcal ener conerted nto mechancal ener as 3 3 cos.3 N.45 m cos. J W = FΔ r θ = = (e) () ( K+ U + U = K+ U ) chemcal B D 3 mb + +. J= md + m( D B) 3 76 (. m s) +. J = 76 D + 76 ( 9.8 m s ) 6.3 m 3 3 ( ) 5.7 J 4.69 J = D = 76 ( K U) ( K U ) + = + D E md + = + m E 5.4 m s where E s the apex o hs moton D ( 5.4 m s) D E D = = =.35 m 9.8 ms

6 () Consder the moton wth constant acceleraton between taeo and touchdown. The tme s the poste root o = + t+ at.34 m = m st+ ( 9.8 m s ) t 4.9t 5.4t.34 = 5.4 ± t = = s 8.44 Δ t = W = Δ K = ( Δ ) m The denst s Δm Δm ρ = = ol A Δ x snce Substtutn ths nto the rst equaton and soln or, Δ x = Δ t, or a constant speed, we et ρa = 3 FIG Also, snce = F, F = ρa Our model predcts the same proportonaltes as the emprcal equaton, and es D = or the dra coecent. Ar actuall slps around the mon object, nstead o accumulatn n ront o t. For ths reason, the dra coecent s not necessarl unt. It s tpcall less than one or a streamlned object and can be reater than one the arlow around the object s complcated = m h = 7.8 m s The retardn orce due to ar resstance s.33. m 3.5 m D A 7.8 m s 38 N = ρ = = Comparn the ener o the car at two ponts alon the hll, or K + U +Δ E= K + U K + U +ΔWe Δ s = K + U where ΔW e s the wor nput rom the enne. Thus,

7 Δ We = Δ s + K K + U U K = K econzn that requred power nput rom the enne as and ddn b the trael tme Δ t es the ΔWe Δs Δ = = + m = + msnθ Δt Δt Δt = 38 N 7.8 m s m s 7.8 m s sn 3. = 33.4 W = 44.8 hp 8.5 m = mass o pumpn = radus o slo top Fr = mar n mcosθ = m When the pumpn rst loses contact wth the surace, n =. = cosθ Thus, at the pont where t leaes the surace. FIG. 8.5 U = Choose n the θ = 9. plane. Then appln conseraton o ener or the pumpn-earth sstem between the startn pont and the pont where the pumpn leaes the surace es K + U = K + U cosθ m + m = + m Usn the result rom the orce analss, ths becomes mcos mcos m θ + θ =, whch reduces to cosθ = 3 θ = cos ( 3) = 48., and es as the anle at whch the pumpn wll lose contact wth the surace (a) Between the second and the thrd pcture, Δ Emech =Δ K+ΔU μmd = m + d ( 5. N m ) d +.5 (. )( 9.8 m s ) d (. )( 3. m s) = [.45 ±.35 ] N d = =.378 m 5. N m

8 Between pcture two and pcture our, Δ Emech =Δ K+ΔU ( d) = m m = 3. m s.45 N.378 m =.3 m s (. ) (c) For the moton rom pcture two to pcture e, Δ Emech =Δ K+ΔU ( D+ d) = (. )( 3. m s) 9. J D = (.378 m ) =.8 m FIG m s 8.55 Δ Emech = Δx E E = d BC μ mh x μ = = md x mh = mdbc BC.38 FIG Let λ represent the mass o each one meter o the chan and T represent the tenson n the chan at the table ede. We mane the ede to act le a rctonless and massless pulle. (a) For the e meters on the table wth moton mpendn, F = + n 5 λ = n= 5λ μ n=.6 5λ = 3λ s s FIG F x = + T s = T = s T 3λ The maxmum alue s barel enouh to support the hann sement accordn to F = + T 3 = T = 3λ so t s at ths pont that the chan starts to slde. Let x represent the arable dstance the chan has slpped snce the start.

9 Then lenth ( 5 x) remans on the table, wth now F = n ( 5 x) λ + = = ( 5 ) λ n x μ.4( 5 ) λ λ.4 λ = n= x = x Consder eneres o the chan-earth sstem at the ntal moment when the chan starts to slp, and a nal moment when x = 5, when the last ln oes oer the brn. Measure hehts aboe the nal poston o the leadn end o the chan. At the moment the nal ln slps o, the = 4 center o the chan s at meters. Ornall, 5 meters o chan s at heht 8 m and the mddle o the danln sement s at 3 8 = 6.5m heht. K + U +Δ Emech = K + U + ( m + m) dx = m + m ( 5λ) 8+ ( 3λ) 6.5 ( λ.4xλ) dx = ( 8λ) + ( 8λ) (.5 m )( 9.8 m s ) dx+ x dx = x 7.5. x +.4 = ( 5.) +.4(.5) = 4..5= 4. = = m s (a) Ener s consered n the swn o the pendulum, and the statonar pe does no wor. So the ball s speed does not chane when the strn hts or leaes the pe, and the ball swns equall hh on both sdes. elate to the pont o suspenson, U = U = m d ( L d), From ths we nd that m( d L) + m = Also or centrpetal moton, m m = where = L d 3L d = 5 Upon soln, we et. θ L e FIG. 8.6 d

10 8.64 (a) At the top o the loop the car and rders are n ree all F = ma m m down = down = Ener o the car-rders-earth sstem s consered between release and top o loop K + U = K + U + mh = m + m ( ) h = + ( ) h=.5.5 Let h now represent the heht o the release pont. At the bottom o the loop we hae mh = m b or b = h F = ma mb nb m= ( up) m( h) nb = m+ At the top o the loop, mh = m t + m ( ) t = h 4 mt F = ma n t m= m nt = m+ h 4 m( h) nt = 5m FIG Then the normal orce at the bottom s larer b ( ) ( ) m h m h nb nt = m + + 5m = 6m