Physics Homewok Solutions Apil 5 though Apil 3, 7 Questions: 6. The oce is pependicula to evey incement o displacement. Theeoe, F =. 6.4 Wok is only done in acceleating the ball om est. The wok is done ove the eective length o the pitche s am the distance his hand moves though windup and until elease. 6. Kinetic enegy is popotional to mass. The ist bullet has twice as much kinetic enegy. 6. No violation. Choose the book as the system. You did wok and the eath did wok on the book. The aveage oce you exeted just countebalanced the weight o the book. The total wok on the book is zeo, and is equal to its oveall change in kinetic enegy. 7.3 Both agee on the change in potential enegy, and the kinetic enegy. They may disagee on the value o gavitational potential enegy, depending on thei choice o a zeo point. 7.5 Lit a book om a low shel to place it on a high shel. The net change in its kinetic enegy is zeo, but the book-eath system inceases in gavitational potential enegy. Stetch a ubbe band to encompass the ends o a ule. It inceases in elastic enegy. Rub you hands togethe o let a peal dit down at constant speed in a bottle o shampoo. Each system (two hands; peal and shampoo) inceases in intenal enegy. 7.8 The oiginal kinetic enegy o the skidding ca can be degaded into kinetic enegy o andom molecula motion in the ties and the oad: it is intenal enegy. I the bakes ae used popely, the same enegy appeas as intenal enegy in the bake shoes and dums. 7. The total enegy o the ball-eath system is conseved. Since the system initially has gavitational enegy mgh and no kinetic enegy, the ball will again have zeo kinetic enegy when it etuns to its oiginal position. Ai esistance will cause the ball to come back to a point slightly below its initial position. On the othe hand, i anyone gives a owad push to the ball anywhee along its path, the demonstato will have to duck.
Poblems: 6.3 Method One. Let φ epesent the instantaneous angle the ope makes with the vetical as it is swinging up om φ i = to φ = 6. In an incemental bit o motion om angle φ to φ + dφ, the deinition o adian measue implies that = ( m )dφ. The angle θ between the incemental displacement and the oce o gavity is θ = 9 +φ. Then cosθ = cos( 9 +φ)= sin φ. The wok done by the gavitational oce on Batman is W = Fcosθd = mg( sin φ) ( m)dφ i = mg m φ =6 6 φ = sin φ dφ 9.8 m s = 8 kg m = ( 784 N )( m )( cos 6 +)= 4.7 3 J 6 ( cosφ) Method Two. The oce o gavity on Batman is mg = ( 8 kg) ( 9.8 m s )= 784 N down. Only his vetical displacement contibutes to the wok gavity does. His oiginal y- coodinate below the tee limb is m. His inal y-coodinate is( m )cos 6 = 6 m. His change in elevation is 6 m m. The wok done by gavity is = 6 m W = F cosθ = ( 784 N )( 6 m)cos8 = 4.7 kj. 6.4 (a) W = mgh = ( 3.35 5 )9.8 ( ) J = 3.8 J Since R = mg, W ai esistance = 3.8 J 6.7 (a) W = F = F x x + F y y = ( 6.) ( 3.) N m + (.) (.) N m = 6. J θ = cos F F = 6 cos 6. + (.) )( 3.) + (.) = 36.9
6. W = F x dx and W equals the aea unde the Foce-Displacement cuve (a) Fo the egion x 5. m, Fo the egion 5. x., ( 5. m ) W = 3. N = 7.5 J W = ( 3. N) ( 5. m)= 5. J Fo the egion. x 5., (d) Fo the egion x 5. ( 5. m ) W = 3. N = 7.5 J W = ( 7.5 + 7.5 + 5.) J = 3. J 6.6 (a) W = F d i.6 m W = ( 5 N + x N m 5 x N m )dx cos 3 x 5 x W = 5 x + 3.6 m W = 9. kj +.8 kj.8 kj = 9. kj Similaly, W = ( 5. kn )(. m )+ W =.7 kj, lage by 9.6% (. kn m ). m ( 5. kn m )(. m) 3 3 6. (a) K A =.6 kg (. m s) =. J mv B = K B : v B = K B m = () 7.5.6 = 5. m s W = K = K B K A = m v ( B v A )= 7.5 J. J = 6.3 J
6.6 (a) K i + W = K = mv + W = ( 5. 3 kg)78 ( m s) = 4.56 kj F = W cosθ = 4.56 3 J = 6.34 kn (.7 m)cos a = v v i x = ( 78 m s) = 4 km s.7 m (d) F = ma= ( 5 3 kg)4 ( 3 m s )= 6.34 kn 6.3 F y = ma y : n + ( 7. N )sin. 47 N = n = 3 N k = µ k n =.3( 3 N ) = 36.9 N (a) W = F cosθ = ( 7. N )( 5. m)cos. = 39 J W = F cosθ = ( 3 N) ( 5. m)cos 9. = J (d) W = F cosθ = ( 47 N )( 5. m)cos 9. = E int = k d = ( 36.9 N )( 5. m)= 85 J (e) K = K K i = W E int = 39 J 85 J = +44 J 6.34 (a) The distance moved upwad in the ist 3. s is y = vt = +.75 m s ( 3. s )=.63 m. The moto and the eath s gavity do wok on the elevato ca: mv i + W moto + mg y cos8 = mv W moto = 65 kg (.75 m s) + ( 65 kg)g(.63 m)=.77 4 J
Also, W = P t so P = W t =.77 4 J = 5.9 3 W = 7.9 hp. 3. s When moving upwad at constant speed v =.75 m s equals the weight = ( 65 kg) ( 9.8 m s )= 6.37 3 N. Theeoe, the applied oce P = Fv = ( 6.37 3 N) (.75 m s)=. 4 W = 4.9 hp. 7. (a) We take the zeo coniguation o system potential enegy with the child at the lowest point o the ac. When the sting is held hoizontal initially, the initial position is. m above the zeo level. Thus, U g = mgy = ( 4 N )(. m)= 8 J. Fom the sketch, we see that at an angle o 3. the child is at a vetical height o (. m )( cos 3. ) above the lowest point o the ac. Thus, U g = mgy = ( 4 N )(. m) ( cos 3. )= 7 J. The zeo level has been selected at the lowest point o the ac. Theeoe, U g = at this location. 7.5(a) : U i + K i = U + K mgh + = mg( R)+ mv g( 3.5R)= g( R)+ v v = 3.gR F = m v R : n + mg = m v R
n = m v R g = m 3.gR g R =.mg n =.( 5. 3 kg)9.8 ( m s ) =.98 N downwad 7.9 Using consevation o enegy o the system o the Eath and the two objects (a) ( 5. kg)g( 4. m)= ( 3. kg)g 4. m + 5. + 3. v v = 9.6 = 4.43 m s Now we apply consevation o enegy o the system o the 3. kg object and the Eath duing the time inteval between the instant when the sting goes slack and the instant at which the 3. kg object eaches its highest position in its ee all. ( 3.)v = mg y = 3.g y y =. m y max = 4. m + y = 5. m 7.5 (a) W OA = dxî yî + x ĵ = ydx and since along this path, y = W OA = W AC = 5. m 5. m 5. m dyĵ yî + x ĵ = x dy 5. m Fo x = 5. m, W AC = 5 J and W OAC = + 5 = 5 J 5. m W OB = dyĵ yî + x ĵ = x dy since along this path, x =, W OB = W BC = 5. m 5. m dxî yî + x ĵ = ydx 5. m since y = 5. m, W BC = 5. J
W OBC = + 5. = 5. J W OC = ( dxî + dyĵ ) ( yî + x ĵ) = ydx + x dy 5. m Since x = y along OC, W OC = ( x + x )dx = 66.7 J (d) F is nonconsevative since the wok done is path dependent. 7.6 Choose the zeo point o gavitational potential enegy o the object-sping-eath system as the coniguation in which the object comes to est. Then because the incline is ictionless, we have : o E B = E A K B + U gb + U sb = K A + U ga + U sa + mg( d + x)sin θ + = + + kx. Solving o d gives d = kx mgsin θ x. 7.3 U i + K i + E mech = U + K : m gh h = m v + m v = µ n = µm g m gh µm gh = ( m + m )v ( hg) v = m µm m + m v = 9.8 ( ms).5 m [ 5. kg.4( 3. kg) ] 8. kg = 3.74 m s 5. m 7.3 (a) W = F x dx = ( x + 4)dx = x + 4x 5. m = 5. +.. 4. = 4. J K + U = U = K = W = 4. J K = K mv K = K + mv = 6.5 J 7.4 (a) Thee is an equilibium point wheeve the gaph o potential enegy is hoizontal:
At =.5 mm and 3. mm, the equilibium is stable. At =.3 mm, the equilibium is unstable. A paticle moving out towad appoaches neutal equilibium. The system enegy E cannot be less than 5.6 J. The paticle is bound i 5.6 J E < J. I the system enegy is 3 J, its potential enegy must be less than o equal to 3 J. Thus, the paticle s position is limited to.6 mm 3.6 mm. (d) K + U = E. Thus, K max = E U min = 3. J ( 5.6 J)=.6 J. (e) Kinetic enegy is a maximum when the potential enegy is a minimum, at =.5 mm. () 3 J+ W = J. Hence, the binding enegy is W = 4 J.