Physics 120. Exam #2. May 23, 2014

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1 Physics 10 Exa # May 3, 014 Nae Please read and follow these instructions carefully: ead all probles carefully before attepting to solve the. Your work ust be legible, and the organization clear. You ust show all work, including correct vector notation. You will not receive full credit for correct answers without adequate explanations. You will not receive full credit if incorrect work or explanations are ixed in with correct work. So erase or cross out anything you don t want graded. Make explanations coplete but brief. Do not write a lot of prose. Include diagras. Show what goes into a calculation, not just the final nuber. For exaple p v " = 5kg ( ) ( s ) =10 kg s Give standard SI units with your results unless specifically asked for a certain unit. Unless specifically asked to derive a result, you ay start with the forulas given on the forula sheet including equations corresponding to the fundaental concepts. Go for partial credit. If you cannot do soe portion of a proble, invent a sybol and/or value for the quantity you can t calculate (explain that you are doing this), and use it to do the rest of the proble. All ultiple choice questions are worth 4 points and each free-response part is worth 8 points Proble #1 /0 Proble # /8 Proble #3 /0 Proble #4 /8 Total /96 I affir that I have carried out y acadeic endeavors with full acadeic honesty.

2 1. ock-clibing is a fun but perhaps very dangerous sport. Suppose that the lead cliber (the leader) has an accident and falls fro a height H above a runner (a fixed etal loop through which the rope runs) while a second (lower) cliber holds fast to liit the fall and does not ove. When the falling leader reaches distance H below the runner, the rope begins to stretch. The stretch is a axiu when the leader has fallen an additional distance d and has coe to a stop. The force on the rope is then a axiu at agnitude F ax. This is the dangerous part of the fall because F ax could be large enough to snap the rope. For any particular rope, the spring constant k depends on the length of rope L and on the elasticity of the rope aterial to stretch e rope, considered to be constant. Thus, for this exaple we can Halliday, esnick, & Walker, Fundaentals of Physics, 7 th Ed. write k = e rope L. a. What is the axiu stretch of the rope, if k =1500 N is 80kg? for this rope if your ass Assue the syste is the rope, person, and the Earth. Fro the energy principle we have, assuing that ΔKE = 0 is zero because the leader starts and ends at rest and taking the zero of the gravitational potential energy at the runner. Thus we have ΔE = 0 = ΔKE + ΔU g = gy f gy i 1 kx i ( ) gh gh gd + 1 kd = 0 ( ) + ( 1 kx f ) = ( g H + d ) + ( 1 kd 0) d = g ± ( g) + 4gkH = 80kg 9.8 s k ( ) ± 80kg 9.8 s ( ) kg 9.8 s 1500 N ( ) = 3 Therefore the axiu stretch is d = 3. b. What is the agnitude of the axiu force? The axiu agnitude of the force is given by Hooke s law: F ax = kd = g + ( g) + 4gkH =1500 N 3 = 4500N

c. Suppose that you have the situation of a short fall. Copared to the axiu force for the longer fall ( F ax,long ), the axiu force for the short fall ( F ax,short ) is 1. greater than F ax,long.

3 c. Suppose that you have the situation of a short fall. Copared to the axiu force for the longer fall ( F ax,long ), the axiu force for the short fall ( F ax,short ) is 1. greater than F ax,long.. less than F ax,long. 3. equal to F ax,long. 4. unable to be deterined fro the inforation given. Halliday, esnick, & Walker, Fundaentals of Physics, 7 th Ed. Since F ax = kd = g + ( g) + 4gkH, we have as L decreases in the expression for k, k increases as does d. Thus the axiu force increases since both d and k. You can also see this using the nubers in the proble and the result of part b.

4 . When one thinks of bones in the huan body, one doesn t norally consider bone as a copressible aterial, but huan bones copress by different aounts when various loads (forces) are applied. Consider for exaple, an average adult ale feur or thighbone. The feur has an average length of approxiately L = 48c and a circular cross-sectional area with a diaeter of d =.3c. When a force is applied over the cross-sectional area of the bone, a stress on the bone is produced and this stress causes the bone to strain under the load. The strain changes the length of the bone of length. We ve used Hooke s law do describe elastic aterials, or aterials that defor under an applied force and when the applied force is reoved return to their original shape. Young s odulus for bone is Y = N. (Data are taken fro Bioedical Applications of Introductory Physics, by J.A. Tuszynski and J.M. Dixon, Wiley, 00 and Clinically Oriented Anatoy, Ed. by K.L. Moore & A.F. Dalley, LWW Publishing, 005) a. Using the generalized expression above for Hooke s law, under noral conditions, by how uch do you copress a single feur when you stand upright at rest? Assue that you have a ass of 60kg. Stress = Y Strain F A = Y ΔL L ΔL = FL AY = F W L AY = gl AY 60kg s ΔL = π.034 = = N Here we are looking at calculating the change in length of a single feur. Each leg supports only one-half of your total weight. b. Suppose that you were to soehow copress your feur by 6 (aybe by falling while rock clibing and by the way this aount is ore than enough to break your feur). What is the ratio of the force needed to break your leg to your weight? Stress = Y Strain F A = Y ΔL L F = YAΔL L N π F = = N = αf W 0.48 α = F = N =141 F W 60kg 9.8 s

5 c. Starting fro rest, through what height would you have to fall landing straightlegged on your feet, so that you could copress a feur by 6? (Hint: ecall Hooke s law fro class and you will need to apply this to the generalized for of Hooke s law above to deterine a value for the stiffness of your feur.) Fro the energy principle we have: ΔE = 0 = ΔKE + ΔU g = 1 v i 0 = g d ( ) g h + 1 k d ( ) ( 1 v f ) + gy f gy i ( ) h = kd g d = N kg 9.8 s ( ) + ( 1 kx f 1 kx i ) ( ) = 0.88 where the effective spring constant of a feur has been deterined fro the generalized for of Hooke s law, Stress = Y Strain F A = Y ΔL L F = AY ΔL = kδl L k = AY L = π N = N Note also that fro proble #1, e rope = AΔy. d. Suppose that instead of landing straight-legged as in the previous part you bend your knees while landing. In this case 1. the work done in bringing you to rest would be the sae, but the force on your feur would be less because your oentu changes over a larger distance.. the work done in bringing you to rest would be the sae, but the force on your feur would be greater because your oentu changes over a saller distance. 3. the work done in bringing you to rest would be the greater, but the force on your feur would be less because your oentu changes over a larger distance. 4. the work done in bringing you to rest would be the saller, but the force would be greater on your feur because your oentu changes over a saller distance. ( ) = FΔy. Dropping fro the The work done is given byw = ΔKE = gh sae height produces the sae change in kinetic energy. But bending your legs akes your oentu change over a uch larger distance. Therefore the force on each of your feurs is less.

6 3. Suppose you have the arrangeent of asses connected to a spring as shown below where 1 = 3kg, = 6kg, k = 10 N, and the angles are θ = The block with ass 1 sits on a horizontal surface while the block of ass sits on the incline. Both portions of the track have friction with coefficient of friction µ = 0.4. k 1 a. At what stretch of the spring will the blocks be oving at a speed of v = 0.8 s? θ There was an error involving the nubers to this question on the exa. The solutions are actually given by solving the quadratic equation for d - which that part is fine. I would expect two positions for which the speed is v = 0.8 s, one when the blocks are speeding up and one when the blocks are slowing down. However with these nubers (ainly the friction part) there are no real solutions. So, everyone gets full credit on this part and the next. b. How uch energy was lost to the surroundings as heat? The energy dissipated as heat is due to friction. The work done by friction would be given if d could have been solved for by ΔE = W fr,1 + W fr, = µf N,1 µf N, = µ 1 gd + gd cosθ ΔE = d 3kg + 6kgcos 30 ( ) = µgd( 1 + cosθ ) ( ) = 3.d

7 c. Suppose that you reove the spring fro the syste above and place it on a horizontal frictionless surface. You attach one end to a wall and to the other end you put ass 3. You then stretch the spring by an aount x ax fro equilibriu (defined to be x eq = 0 ). When you release the ass fro rest, the kinetic and potential energies change according to which of the following graph? (Key: ed = KE and Blue = U s ) 1.. E ax KE,U S E ax KE,U S -x ax x ax -x ax x ax E ax E ax KE,U S KE,U S -x ax x ax -x ax x ax Since there s no friction, at any point the su of the kinetic and potential energies have to su to the total energy by: ΔE = 0 = ΔKe + ΔU g = 1 v 0 1 kx ax = E ax = 1 v + 1 kx ( ) + ( 1 kx 1 kx ax )

8 4. An auseent park thrill ride consists of a cart with soe riders (of total ass 500kg ) that are set in otion by a large spring with spring constant 800 N. The cart and riders travels along the flat horizontal section of track that is located 15 above the ground and then down the rap toward the loop-the-loop, which has an unknown diaeter. The entire track is frictionless unless otherwise stated. a. What will be the speed of the cart and riders half way down the rap if the spring were initially copressed by 3 and there were a friction present only for the launch portion of the ride? (Assue µ = 0. ) Using the energy principle assuing that the syste is the cart, riders, and the earth we have since the only external force is friction ΔE = W fr = ΔKE + ΔU g + ΔE rest = 1 v f 1 v i ( ) + ( gy f gy i ) + 1 kx f 1 ( kx i ) µf N = µgx i = 1 kx i + 1 v f + gy f gy i v f = ( ( ) + 1 kx i µgx i ) g y i y f v f = 500kg 9.8 s ( ( ) N ( 3.0) ) 500kg = 11 s b. If the speed of the cart and riders at the top of the loop were 9 s, what is the radius of the loop? Applying the energy principle between the point the cart and riders are half-way down the incline and the point at the top of the loop we have assuing the syste is the cart, riders and the earth ΔE = 0 = ΔKE + ΔU g + ΔE rest = 1 v t 1 v a y t = y a + g 1 v ( a v t ) = radius = y t = s 11 s ( ) + gy t gy a (( ) ( 9 s ) ) = 9.5 ( )

9 c. How uch work did the gravitational force on the cart and riders? The work done is the difference in the kinetic energies and we have W = ΔKE = ΔU = 1 v t v b ( ) = 1 500kg 9 s (( ) ( 16.4 s ) ) J = 4.70kJ where the speed at the botto of the loop is calculated using the energy principle and is ΔE = 0 = ΔKE + ΔU g + ΔE rest = 1 v b 1 v a ( ) + gy b gy a v b = v a + gy a = ( 11 s ) = 16.4 s s ( ) Or we could explicitly calculate the work done by gravity. W = F net d r y f y f = r = 0, g,0 dx,dy,dz g dy y i W = gδy = 500kg 9.8 s 9.5 = 4.7kJ y i=0 d. Which of the following expressions gives the noral force on the cart and riders at the top of the loop? 1. g + v. g v g + v 5. g v 6. g v Fro the oentu principle, we have F net = d p dt 0, g F N,0 = 0, v F N = g v

10 Physics 10 Equations r =< r x,r y,r z >= r ˆr agnitude of a vector :r = r = r x + r y + r z r unit vector : ˆr = r v = Δ r Δt ; v v avg = i + v f r f = r i + v avg Δt F net = Δ p Δt p = γ v; li p v<<c ( ) ~ v p f = p i + F net Δt r f = r i + v F i Δt + net ( Δt) F G = GM 1M ˆr r F g ~ g Constants: g = 9.8 s G = N kg e = kg = 0.51 MeV c p = kg = MeV c E = kg E = N A = c = s 1eV = J ax + bx + c = 0 x = b ± b 4ac a d p dt = F net F s = k s F fr = µf N k parallel = 1 k series = ρ = V Y = k IAB d N N i=1 k i 1 i=1 k i Stress = Y Strain; Stress = F Δl ; Strain = A l F " = dp dt ˆp F = p dˆp dt F F Net = F " + F = d p = v ; dt = dp dt ˆp + p dˆp dt ΔE = ΔQ + ΔW = ΔKE + ΔU g + ΔE rest ΔW = ΔKE = F d r ; F d r = F x dx + F y dy + F Z dz KE = ( γ 1)c ; li( KE) ~ 1 v v<<c E T = γ c ; E T = p c + c 4 ; E rest = c U g = gy U s = 1 kx T = π k Inclines: F N = gcosθ; dv dt = gsinθ

Physics 140 D100 Midterm Exam 2 Solutions 2017 Nov 10

Physics 140 D100 Midterm Exam 2 Solutions 2017 Nov 10 There are 10 ultiple choice questions. Select the correct answer for each one and ark it on the bubble for on the cover sheet. Each question has only one correct answer. (2 arks each) 1. An inertial reference