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Unformatted text preview: alameri (aoa434) – Hwk07 – Stokes – (19102)1This printout should have 21 questions.Multiplechoice questions may continue onthe next column or page – find all choicesbefore answering.This is ONLINE HOMEWORK No 7. It isdue before 7.00 am on MONDAY 16 March,Abu Dhabi Time (9.00 pm on Sunday inTexas)001(part 1 of 3) 10.0 pointsA single conservative forceF(x) =bx+aacts on a 4.5 kg particle, wherexis in meters,b= 3.23 N/m anda= 3.06 N.As the particle moves along thexaxis fromx1= 1.06 m tox2= 4.58 m, calculate thework done by this force.Correct answer: 42.8335 J.Explanation:The work done by the conservative force isW=integraldisplayFxdx=integraldisplay(bx+a)dx=12bx2+axvextendsinglevextendsinglevextendsinglex2x1=12(3.23 N/m)x2+ (3.06 N)xvextendsinglevextendsinglevextendsingle4.58 m1.06 m= 47.8917 J5.05821 J= 42.8335 J.002(part 2 of 3) 10.0 pointsCalculate the change in the potential energyof the particle.Correct answer:42.8335 J.Explanation:From conservation of energyΔK+ ΔU= 0,we obtainΔU=ΔK=W=42.8335 J.003(part 3 of 3) 10.0 pointsCalculate the particle’s initial kinetic energyatx1if its final speed atx2is 16.3 m/s.Correct answer: 554.969 J.Explanation:The change in the kinetic energy isΔK=KfKiKi=KfΔKKi=12mv2fΔK=12(4.5 kg)(16.3 m/s)242.8335 J= 554.969 J.004(part 1 of 2) 10.0 pointsA 0.2 kg bead slides on a curved wire, startingfrom rest at point A as shown in the figure.The segment from A to B is frictionless, andthe segment from B to C is rough. The pointA is at height 8.9 m and the point C is atheight 0.8 m with respect to point B.The acceleration of gravity is 9.8 m/s2.A8.9 mB.8 mCFind the speed of the bead at B.Correct answer: 13.2076 m/s.Explanation:Given :m= 0.2 kgandh= 8.9 m.Choose the zero level for potential energy atthe level of B. Between A and B,KA+UA=KB+UB0 +mg H=1wmv2+ 0v=radicalbig2g H=radicalBig2(9.8 m/s2)(8.9 m)= 13.2076 m/s.alameri (aoa434) – Hwk07 – Stokes – (19102)2005(part 2 of 2) 10.0 pointsIf the bead comes to rest at C, find the changein mechanical energy due to friction as itmoves from B to C.Correct answer:15.876 J.Explanation:Given :h= 0.8 mChoose the starting point at B, the zero potential energy level at B, as before, and theend point at C. Then the energy loss is equalto the work done by the nonconservativeforces and isWnc=KfKi+UfUi= 012mv2B+mg h=(0.2 kg) (13.2076 m/s)22+ (0.2 kg) (9.8 m/s2) (0.8 m)=15.876 J.006(part 1 of 3) 10.0 pointsA block starts at rest and slides down a frictionless track except for a small rough area ona horizontal section of the track (as shown inthe figure below).It leaves the track horizontally, flies throughthe air, and subsequently strikes the ground.The acceleration of gravity is 9.81 m/s2....
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This note was uploaded on 02/16/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
 Spring '09
 NAJAFZADEH
 mechanics, Work

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