# hw5 - al-ameri(aoa434 – Hwk05 – Stokes –(19102)1This...

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Unformatted text preview: al-ameri (aoa434) – Hwk05 – Stokes – (19102)1This print-out should have 20 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.This is ONLINE HOMEWORK No 5. It isdue before 7.00 am on Tuesday 24 February,Abu Dhabi Time (9.00 pm on Monday inTexas)00110.0 pointsHint:This problem requires a train of logic.(1) Analyze force diagram,(2) use Newton’s Laws, and(3) solve the equations of motion.A block starts from rest at a height of 3.9 mon a fixed inclined plane.The acceleration of gravity is 9.8 m/s2.2.8kgμ=.1334◦What is the speed of the block at the bot-tom of the ramp?Correct answer: 7.85541 m/s.Explanation:Let :h= 3.9 m,m= 2.8 kg,μ= 0.13,θ= 34◦,andvf= final speed.FfNmg34◦dh34◦The normal force to the inclined plane isN=mgcosθ. The sum of the forces par-allel to the inclined plane isFnet=ma=mgsinθ-μ mgcosθa=gsinθ-μ gcosθSincev2f=v2+ 2a x= 2a d(1)along the plane and the distance moved alongthe plane isd=hsinθ(2)thereforevf=radicalbigg2a hsinθ=radicalbigg2g h(sinθ-μcosθ)sinθ=radicalbig2g h(1-μcotθ)(3)=radicalBig2 (9.8 m/s2) (3.9 m) [1-(0.13) cot34◦]= 7.85541 m/s.00210.0 pointsA 77 kg person escapes from a burning build-ing by jumping from a window 21 m above acatching net.The acceleration of gravity is 9.81 m/s2.al-ameri (aoa434) – Hwk05 – Stokes – (19102)2Assuming that air resistance is simply aconstant 85 N force on the person during thefall, determine the person’s velocity just be-fore hitting the net.Correct answer:-19.1221 m/s.Explanation:Basic Concepts:Fnet=ma=FR-Fg=FR-mgv2f= 2aΔysincevi= 0 m/s.Given:m= 77 kgΔy=-21 mFR= 85 Ng= 9.81 m/s2Solution:85 NaFgNote:Figure is not drawn to scale.a=FR-mgm=85 N-(77 kg)(9.81 m/s2)77 kg=-8.7061 m/s2vf=radicalBig2 (-8.7061 m/s2)(-21 m)=±19.1221 m/sThe velocity is-19.1221 m/s since the direc-tion is down.00310.0 pointsA fire helicopter carries a 640 kg bucket at theend of a 9.8 m long cable. When the helicopteris returning from a fire at a constant speed of18 m/s, the cable makes an angle of 21.1◦withrespect to the vertical.The acceleration of gravity is 9.8 m/s2.21.1◦Find the horizontal force exerted by airresistance on the bucket.Correct answer: 2420.16 N.Explanation:Given :m= 640 kg,θ= 21.1◦,g= 9.8 m/s2,andv= 18 m/s.TRmgThe bucket is in equilibrium, soΣFy=Tcosθ-mg= 0Tcosθ=mgΣFx=Tsinθ-R= 0Tsinθ=RDividing,TsinθTcosθ=RmgR=mgtanθ= (640 kg)(9.8 m/s2)(tan21.1◦)=2420.16 N.00410.0 pointsThe static friction between the two blocks isal-ameri (aoa434) – Hwk05 – Stokes – (19102)3.61 and kinetic friction between the blockwith mass 7.9 kg and the horizontal surface is.3.The acceleration of gravity is 9.8 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.

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hw5 - al-ameri(aoa434 – Hwk05 – Stokes –(19102)1This...

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