hw6 - al-ameri (aoa434) – Hwk06 – Stokes –...

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Unformatted text preview: al-ameri (aoa434) – Hwk06 – Stokes – (19102)1This print-out should have 22 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.This is ONLINE HOMEWORK No 6. It isdue before 7.00 am on Wednesday 3 March,Abu Dhabi Time (9.00 pm on Tuesday inTexas).00110.0 pointsThe initial kinetic energy imparted to a.51 kg bullet is 1859 J.The acceleration of gravity is 9.81 m/s2.Neglecting air resistance, find the range ofthis projectile when it is fired at an angle suchthat the range equals the maximum heightattained.Correct answer: 0.348882 km.Explanation:Let :m= 0.51 kg,K= 1859 J,andg= 9.81 m/s2.Assume that the firing height is negligible andthat the bullet lands at the same elevationfrom which it was fired. The rangeRis givenbyR=v2gsin 2θ .From kinematics,x= (vcosθ)tandy= (vsinθ)t−12g t2.R=hmax=(vsinθ)22g, sov2gsin 2θ=(vsinθ)22gv2g(2 sinθcosθ) =(vsinθ)22gtanθ= 4θ= 76◦.The kinetic energy isK=12mv2v2=2Km,soR=v2gsin2θ=2Kmgsin 2θ=2 (1859 J)(0.51 kg) (9.81 m/s2)sin 152◦·km1000 m=.348882 km.002(part 1 of 3) 10.0 pointsA 18.7 kg block is dragged over a rough, hor-izontal surface by a constant force of 85.8 Nacting at an angle of 28◦above the horizon-tal. The block is displaced 50.9 m, and thecoefficient of kinetic friction is 0.19.The acceleration of gravity is 9.8 m/s2.18.7 kgμ= 0.1985.8N28◦Find the work done by the 85.8 N force.Correct answer: 3856.03 J.Explanation:Let :m= 18.7 kg,F= 85.8 N,sx= 50.9 m,μ= 0.19,andg= 9.8 m/s2.Consider the force diagramal-ameri (aoa434) – Hwk06 – Stokes – (19102)2FθmgnfkWork isW=vectorF·vectors, wherevectorsis the distancetraveled. In this problemvectors= 5ˆıis only in thexdirection.⇒WF=Fxsx=F sxcosθ= (85.8 N) (50.9 m) cos28◦=3856.03 J.003(part 2 of 3) 10.0 pointsFind the work done by the force of friction.Correct answer:−1382.75 J.Explanation:To find the frictional force,Ffriction=μN,we need to findNfrom vertical force balance.Note thatNis in the same direction as theycomponent ofFand opposite the force ofgravity. ThusFsinθ+N=mgso thatN=mg−Fsinθ .Thus the friction force isvectorFfriction=−μNˆı=−μ(mg−Fsinθ)ˆı.The work done by friction is thenWμ=vectorFfriction·vectors=−|fμ||s|=−μ(mg−fμsinθ)sx=−(0.19) [(18.7 kg) (9.8 m/s2)−(85.8 N) sin 28◦] (50.9 m)=−1382.75 J.004(part 3 of 3) 10.0 pointsIf the block was originally at rest, determineits final speed.Correct answer: 16.2641 m/s.Explanation:The net work done on the block is equal tothe change in kinetic energy, so12mv2−0 =WF+Wμv=radicalbigg2 [WF+Wμ]m=radicaltpradicalvertexradicalvertexradicalbt2bracketleftBig(3856.03 J) + (−1382.75 J)bracketrightBig(18.7 kg)=16.2641 m/s....
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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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hw6 - al-ameri (aoa434) – Hwk06 – Stokes –...

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