# hw4 - al-ameri(aoa434 – Hwk04 – Stokes –(19102)1This...

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Unformatted text preview: al-ameri (aoa434) – Hwk04 – Stokes – (19102)1This print-out should have 19 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.This is Online Homework No 4. It is duebefore 7.00 am on Tuesday 17 February (9.00pm on Monday in Texas)00110.0 pointsAn elevator is being lifted up an elevator shaftat a constant speed by a steel cable as shownin the figure below. All frictional effects arenegligible.steelcableElevator going upat constant speedIn this situation, forces on the elevator aresuch that1.the upward force by the cable is greaterthan the downward force of gravity.2.the upward force by the cable is equal tothe downward force of gravity.correct3.None of these.(The elevator goes upbecause the cable is being shortened, not be-cause an upward force is exerted on the eleva-tor by the cable.)4.the upward force by the cable is smallerthan the downward force of gravity.5.the upward force by the cable is greaterthan the sum of the downward force of gravityand a downward force due to the air.Explanation:Since the elevator is being lifted at a con-stant speed, the net force on it is zero, there-fore, the upward force by the cable is equal tothe downward force of gravity.00210.0 pointsA toy car is given a quick push so that it rollsup an inclined ramp. After it is released, itrolls up, reaches its highest point and rollsback down again.Friction is so small it canbe ignored.Whatnet forceacts on the car?1.Netconstantforcedownthe rampcor-rect2.Netdecreasingforcedownthe ramp3.Netincreasingforceupthe ramp4.Net force ofzero5.Netconstantforceupthe ramp6.Netincreasingforcedownthe ramp7.Netdecreasingforceupthe rampExplanation:The net force is the component of gravityalong the incline; it is constant and acts downthe ramp.003(part 1 of 2) 10.0 pointsA 29.6 kg mass attached to a spring scale restson a smooth, horizontal surface. The springscale, attached to the front end of a boxcar,readsT= 49.6 N when the car is in motion.mIf the spring scale reads zero when the caris at rest, determine the acceleration of thecar, when it is in motion as indicated above.Correct answer: 1.67568 m/s2.al-ameri (aoa434) – Hwk04 – Stokes – (19102)2Explanation:Basic Concept:Fictitious forcesInertial/non-inertial framesSolution:In the laboratory frame (which is aninertial frame):The block is accelerating with the car, withan acceleration ofa. This requires an externalforce which must equalm a. Since the tensionTis theONLYavailable, external force weknow thatm a=T,so thata=Tm=49.6 N29.6 kg= 1.67568 m/s2Alternative derivation in the acceler-ated frame of the car:The acceleration of the mass in the refer-ence frame of the car isa′= 0; on the otherhand, the forces acting on the mass are thetensionTand the fictitious forceFfic=ma,which acts in the direction opposite to thecar’s motion. Thus,T−Ffic=T−ma=ma′= 0and we geta=Tm= 1.67568 m/s2004(part 2 of 2) 10.0 pointsWhat would be the reading on the scale if theboxcar were moving at a constant velocity?boxcar were moving at a constant velocity?...
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hw4 - al-ameri(aoa434 – Hwk04 – Stokes –(19102)1This...

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