hw06 - Physics 7a Spring 2008 Homework 6 Solutions Problem...

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Unformatted text preview: Physics 7a Spring 2008 Homework 6 Solutions Problem 1: 8.27 An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable breaks when the elevator is at a height habove the top of the spring, calculate the value that the spring constant éizshould have so that passengers undergo an acceleration of no more than 7.0 g when brought to rest. Let iiifbe the total mass of the elevator and passengers. Solution Steps: 1) The total energy before the elevator drops (Mgh) must equal the total energy when the spring is fully compressed (1/2 k x2 — mgx) Notice that the final energy includes a loss in gravitational potential energy equal to mgx 2) Draw a free body diagram of the elevator with the compressed spring to show that the maximum compression xmax < 8mg/k 3) plug in your x value in step 2 into your energy equation in step 1 4) Solve for k Physics 7a.Spring 2.008 Homework 6 Solutions Problem 2: 8.28 A skier of mass mstarts from rest at the top of a solid sphere of radius i’and slides down its frictionless surface. A) At what angle lsi(see the figure) will the skier leave the sphere? «£2 Solution Steps: 1) Draw a free body diagram and determine the sum of the forces in the x and y direction when the skier is at a small angle 9 2) The sum of the forces in the y direction (toward the center of the Sphere) results in a centripetal acceleration. 3) Use Energy Conservation to determine the velocity as a function of the angle 6 (based on the change in height) 4) The skier leaves the sphere when the normal force is zero 5) Determine 9. '9 N ‘ 2) EFJC : {’13 Sing; 5/ ii "“6” A.” n A Ti} ES}: me33 p m w“ 9%“ _‘ ' = last-101038) "a l\l filing-{1.1536 2 -— VHF/Zara 10 When M : a yr the) Cofig anvil/(2 9) (203(9: 2(r--cose) gceséf; :25 1210*st l 3C039'19~_ alt/”Hi? : {49.51“ . .__._____________ B) If friction were present, would the skier fly off at a greater or lesser angle? {9/ 13mm“: ““34”” : fimwl r NY :> wetecaLtr__1's glomv VJ, F’N‘c-{dl‘rh L 7;? from NW) :1 in 3R? '4) above: COS 61' "" iQ (:68 & TS smq {Lav Mn % (Hut/3i; I be. baj j-Qv W Physics 7a Spring 2008 Homework 6 Solutions Problem 3: 8.42 A spring (R) has an equilibrium length of ((5). The spring is compressed to a length of (112 E) and a mass‘(m,)is placed at its free end on a frictionless slope which makes an angle (9) with respect to the horizontal. The spring is then released. A) If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest? Solution Steps: 1) Draw the mass in the various positions with the spring compressed or at equilibrium 2) The stored energy in the spring is transferred to gravitational potential energy in the form of ' mgh (relative to the starting point) mgh—m gd(51n8) 3) 3The distance 1s the diagonal distance traveled by the mass Physics 7a Spring 2008 Homework 6 Solutions Problem 3: 8.42 (continued) B) If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest? (Na? WWW +0 Gaels) Solution Steps: 1) Draw the mass in the various positions with the spring compressed, at equilibrium or stretched. It helps to draw all three positions in line with each other with the compression distance (x1) and the stretched distance (X2) clearly labeled 2) The stored energy in the compressed spring (I/2k[x1]2) transfers to stored energy in the stretch spring (J /2k[x;] 2) + the gain in gravitational potential energy (mg(x1+x2)sin8) 3) The distance is the diagonal distance traveled by the mass (X1+Xz) Physics 7a Spring 2008 Homework 6 Solutions Problem 3: 8.42 (continued) C) Now the incline has a coefficient of kinetic friction 5:? If the block, attached to the spring; is observed to stop just as it reaches the spring‘s equilibrium position, what is the coefficient of friction W? xeihl VE/él Solution Steps: 1) Draw the mass in the various positions with the spring compressed or at equilibrium. 2) The stored energy in the compressed spring (1/2k[I/2£]2) transfers a loss in heat (equal to the work done by friction (F fl,“ [1126]) + the gain in gravitational potential energy (mg(x1+xz)sin9) 3) Solve for u 2 l»): WE + 3‘ on 2‘ _L 2. Physics ?'a Spring 2008 Homework 6 Solutions Problem 4: 8.44 Early test flights for the space shuttle used a "glider” (mass of m including pilot). After a horizontal launch at v; at a height of h, the glider eventually landed at a speed of vf. A) What would its landing speed have been in the absence of air resistance? Solution Steps: 1) Draw the plane and its path from on top of the hill. 2) The initial Kinetic Energy (1/2m[v;]2) at the top + the Gravitational Potential Energy at the top (mgh) is transferred to Kinetic energy at the bottom (1!2m[vf]2) 3) Solve for v; t 2 k—erjz“ EEWVE “”5 F 2 3" C? were. prof ‘Mw’ll Mia: lg—BG $13: B) What was the average force of air resistance exerted on it if it came in at a constant glide angle of 14 I. to the Earth‘s surface? Solution Steps: 1) The initial Kinetic Energy (1f2m[v,] 2) at the to]: + the Gravitational Potential Energy at the top (mgh) 1s transferred to Kinetic ene1gy at the bottom (1f2m[vfi i)+ a loss to friction forces (W19: Ff" hsinB) 2) Solve for Ff, igwma: W+ w .— 1131: :mm 151‘ “will Physics 7a Spring 2008 Homework 6 Solutions Problem 5: 8.74 A bicyclist coasts down a hill at angle (8) at a steady speed (v) A) Assuming a total mass (m) (bicycle plus rider), what must be the cyclist's power output to climb the same hill at the same speed? Solution Steps: 1) Draw a free body diagram of the bicyclist going down the hill. 2) Show that the force of friction (Ff,) traveling at a constant speed is equal to the downhill component of the weight (mgsinB) (constant v means no acceleration) 3) In a separate free body diagram of the bicyclist traveling uphill (the friction force from wind and internal friction of the bike is the same up and downhill since the speed is the same) show that the force needed by the rider is equal to the sum of friction and the uphill component of weight ' 4) Power is Worldtime. That is the same as: (Force x displacement)! time = Force X velocity Pawns ~‘ (:42 :lng 5:019 ”vi Physics 721 Spring 2008 Homework 6 Solutions Problem 6: 8.77 The potential energy of the two atoms in a diatomic (two-atom) molecule can be written . :1 (J .. Uttlwwgsflsé ' 3 where ris the distance between the two atoms and l:i’ancl 'F’are positive constants. - __ if? , , _ A) At what values of 1F“is é Ea mlmmum? A max1mum? When derivative = 0 (kw) «- ~ i iLI):C:o~(7*\7—ior350 ctr G; a . _.——. 1-_ 12b (7 Ta {9 0k " 1%9 . if “tram-E} B) Atwhat values of'r‘is {I} ? i ' 3 Jr 1:; [[6 ?r\._ EL - J—L (9 W“ _ “k b ._ Ck'" it, ‘( t: (gay w r me C) Describe the motion of one atom with respect to the second atom when E {- ”, Physics "is Spring 20138 Homework 6 Solutions DJ Let f be the fofee one atom exerts on the other. For what values of I'is F 2’ 0? Ftddj}: :‘émr-7 1“ i213 rd? -_--.:_7 O "'K' *éir"j 'i*-I"2int-"iiiI res ~___fi______‘___ E) For whatvslnee oi‘"is ‘L r: H? _. {a (9’5) <1” 4.43 F} For what values of I'is P 3? LJIWEIA. i" : (2—5)er (3] Determine Pas a function of ". Siam“. {Lg enLewa. ‘ thsies l'a Spring 20% Homework t5 Solutions Problem '3": 9.3 Air traveling at veloeitjvr {v} strikes head-on the face of a building of width {w} and height [h] and is brought to rest. . k . . if air has a mates of 1.3 finer eubie meterj determine the average foree of the wind on the building. i. '4' T t I;- F t t r. 1' It_ Solution Steps. The ice}.r here 1s that Foree is the derivative of momentum The rest is messing around with the equations for densityr and flow rate shown below: - l: :r if. alErisii-II : El“. 1: .1; V? Volume Vtioci'i-lr Physics Te Spring 2mg Homework 6 Solutions Problem 8: 9.13 A child in a boat throws a 4.45“ 1"” package out horizontally with a speed of loo "1’“. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 24.3 kg and that of the boat is 35.0 kg. {Take the package's direction of motion as positive.) Solution Steps: The key here is that the conservation of momentum requires that file momentum of the system {childi’hoet and box] is still zero after the ho}: is thrown as there are no significant external forces acting on the system. f...“ if“! ‘5} :1— M3 U f3 L H I .4“— .4“: in all {‘2 + E. ml} \_.-' e songw- Ti’u. igafit +1NiniiS 675's}: 3%“ ills-— like nflfia‘ftv—QL I: oil.'reu;~Li“aim ' Physics Ta Spring Zilflli Homework IS Solutions Problem 9: 9.21 A projectile with mass (3m) is fired with a speed of (up) at an angle (El), breaks into three pieces of equal mass {m} at the highest point of its arc {where its velocityr is horizontal). Two of the fragments move with the same speed right after the explosion as the entire projectile had just before the explosion; one of theae moves verticallyr downward and the lother horizontallv. A} Determine the magnitude of the velocity of the third fragment immediater after the explosion. ""b'm-‘JUEUSG _’ Solution Steps: 1} The velocityr of the projectile prior to explosion is the horizontal component of vn {vncanl} and it is given that that same velocity is the speed of the mini projectiles l and 2 alter the explosion (in the ii and "j directions} 2} Momentum consenration requires that the momentum before the explosion [3mm] is the same after the explosion {sum olimlvl + mgvg + mgvg} 3} Solve for the i andj components of v3. --~‘ Willow + 3 "We L0“: ‘9 J ”Va Cos 6 4,. "V3 5‘ _._,.--'——___,.--—'——____,__\_H-“‘_ affil- - :21!“ CL"; [93‘ y’lwseeég:lfzfiaLaSfiJ1+é/lflgl9)___ "j: 31; rgwjlé}. :(To éfiiflg fl Physics Ta Spring ZUUS Homework 6 Solutions B) Determine the direction of the 1.retocity of the third fragment immediately after the explosion. Soittttort Steps: 1] Draw the vector 1:3 as a sum of its i and j components {heaci to tail} 2) Use your favorite trig relation (1 use tangent cause it is awesome] 3] Solve for El ’fidfifl'fi E}- i} 2"!” coSEr C) Determine the energy released in the explosion. Solution Steps: 1) The Kinetic Energy of the projectile before the explosion is: anywhere)“: 2} The Kinetic Energy of tlte projectile alter the explosion is: 1 ”air" {ft/gross?) Jr émfivymsoj + 1"— m{‘h3£o§(9’ Big) 1 first (2 i- 5 )(Vflz cot-1(3) 3} The change in KE shows more energy otter the explosion (negative flKE} This indicates t1 release of stored chemical energy during the explosion AM: K5,}. - he; : gamma-=51” - gfirvizeoseefl 2 (”(51 (7*.2.) Mvszcoslfi I; iUU'iQ 1th = mots; ”if 4%. SIM-fill pieces. In “+145 cflifmhfl‘4-Ikflh ...
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