HW05-solutions

# HW05-solutions - baum(lmb2768 HW05 distler(56295 This...

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baum (lmb2768) – HW05 – distler – (56295) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. There are various ways to do part 2 oF the ”Wind and Pendulum” problem. The easiest is to fnd the angle at which the three Forces on the ball (tension, gravity and ±) add up to zero. In part 1 oF ”Pulling Two Masses”, you are asked about the ”work done against Fric- tion”. This is very Funny phraseology. The Friction does negative work, and what you are being asked-For is the magnitude (a positive number) oF the work done by Friction. 001 10.0 points Ab lockslidingonahorizontalsurFacehasan initial speed oF 0.5 m/s. The block travels a distance oF 1 m as it slows to a stop. What distance would the block have trav- eled iF its initial speed had been 1 m/s? 1. 1m 2. 2m 3. 3m 4. 4m correct 5. more inFormation is needed to answer the question 6. 0.5 m Explanation: W nc = - f k L =0 - 1 2 mv 0 2 - μmgL = - 1 2 0 2 L = v 0 2 2 μg So the distance is proportional to the square oF the initial speed. 002 (part 1 of 2) 10.0 points Aba l lhav ingma s s1 . 1kgi sconn e c t edby astr ingo Flength0 . 8mtoap ivotpo intand held in place in a vertical position. A constant wind Force oF magnitude 17 . 85 N blows From leFt to right. Pivot Pivot FF (a) (b) H m m L L IF the mass is released From the vertical po- sition, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The accelera- tion oF gravity is 9 . 8m / s 2 . Correct answer: 1 . 1724 m. Explanation: Let : m =1 . 1kg , L . , F =17 . 85 N , and g =9 . / s 2 . Take the original point to be the one where the ball is released and the fnal point where its upward swing stops at height H .Thenthe horizontal displacement is x = ± L 2 - ( L - H ) 2 Since the Force is purely horizontal, it does work W = ² F · d s = F ³ 2 LH - H 2 . Applying the work-energy theorem, ( K + U ) i + W wind =( K + U ) f 0+0+ F ³ 2 LH - H 2 =0+ mgH. H = 2 L 1+ ´ mg F µ 2 · F 2 F 2 = 2 LF 2 F 2 +( mg ) 2

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baum (lmb2768) – HW05 – distler – (56295) 2 = 2(0 . 8m)(17 . 85 N) 2 (17 . 85 N) 2 +(1 . 1kg) 2 (9 . 8m / s 2 ) 2 = 1 . 1724 m . 003 (part 2 of 2) 10.0 points What will be the equilibrium height of the mass? Correct answer: 0 . 38643 m. Explanation: Considering components of the force equi- librium ± F =0, - T sin θ + F =0 and T cos θ - mg =0 . Dividing these, tan θ = F θ =arctan ² F ³ ´ 17 . 85 N (1 . 1kg)(9 . / s 2 ) µ =58 . 8713 , so the equilibrium height is h = L (1 - cos θ ) =(0 . 8m)(1 - cos 58 . 8713 ) = 0 . 38643 m . 004 10.0 points ANavyjetofmass10kglandsonana ircraft carrier and snags a cable to slow it down. If the cable acts like a spring that has a spring constant of 160 N / mandtheaircraftis stopped by this spring in 25 m after hooking onto the cable, what is the speed of the jet when it Frst hooks on to the cable (assume the jet does no breaking on its own)? 1. 70 m/s 2. 100 m/s correct 3. 50 m/s 4. 150 m/s 5. 20 m/s Explanation: Let : m =10kg , k =160N / m , and x =25m Energy is conserved: K i + U i = K f + U f The initial kinetic energy is K i = 1 2 mv 2 i , where v i is the speed of the jet when it Frst hooks on to the cable. The initial potential energy stored in the cable is U i since it is initially unstretched.
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HW05-solutions - baum(lmb2768 HW05 distler(56295 This...

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