1Mon. Feb. 12 Phy107 Spr07 Lecture 101Hour Exam #1• Hour Exam I, Wed. Feb. 14, in-class (50 minutes)• Material Covered: Chap 1, 3-6• One page of notes (8.5” x 11”) allowed• 20 multiple choice questions• Scantron sheets will be used -bring #2 HB pencils and calculator• In-class review todayOn-line review questions, previous exams, at course web siteuw.physics.wisc.edu/~rzchowski/phy107Mon. Feb. 12 Phy107 Spr07 Lecture 102Power ! P =Worktime, Joules(J)second(s) " Watts (W)Power is the rate at which work is doneIt is measured in Watts.(also Horsepower, 1 horsepower = 750 Watts)Mon. Feb. 12 Phy107 Spr07 Lecture 103QuestionYou run up the first flight of stairs, then walkup the second flight. How do the work andpower compare on the two flights?A. Work same, power sameB. Work different, power sameC. Work different, power differentD. Work same, power differentMon. Feb. 12 Phy107 Spr07 Lecture 104QuestionYou press a 500 N weight up to arms length(0.8m) in 2 seconds. Your power outputwhile lifting wasA. 400 JB. 400 WC. 100 WD. 200 WPower = work / time = Force x Distance / time = (500N)x(0.8m)/2 sec = 400 J / 2 sec = 200 WattsMon. Feb. 12 Phy107 Spr07 Lecture 105QuestionYou weigh 500 N and run up a stairs with a 4 mvertical height. You do this in 5 seconds. Yourpower output isA. 400 WB. 200 WC. 100 WD. 50 WWeight = 500 N = mgEnergy change = (mg)h = 2,000 JPower = energy / time = 400 WMon. Feb. 12 Phy107 Spr07 Lecture 106Chapter 1: Post-Aristotle• Inertia:A body subject to no external forces will– Stay at rest if it began at rest– Will continue motion in straight line atunchanging speed if it began in motion.• Can explain how moon keeps orbiting aroundearth, etc.• But need details to quantify this2Mon. Feb. 12 Phy107 Spr07 Lecture 107Speed and acceleration ! Average speed = distance traveledtraveling timeAs an equation: ! Distance traveled = dTraveling time = tAverage speed = s ! s =dt ! Acceleration is the rate at which velocity changes :Acceleration = change in velocitytime to make the changeInstantaneous speed and accel = average valuesover short time interval.Mon. Feb. 12 Phy107 Spr07 Lecture 108VelocityWhich statement best describes the velocity of thecar?A. Increasing with timeB. Decreasing with timeC. Always zeroD. Initially increasing, then decreasingE. Initially decreasing, then increasing.timepositionMon. Feb. 12 Phy107 Spr07 Lecture 109Question: accelerationYou throw a ball directly upwards to the ceilingand let it hit to the floor.Using g=10 m/s2, the acceleration is largestA. Near the ceilingB. Just before it hits floorC. Just after it leaves my handD. None of the aboveAfter it leaves your hand, acceleration is constant, andequals acceleration of gravity. Acceleration is differentthan velocity. The velocity is zero at the top of the arc, butit is still continuously changing, even when it is zero.Mon. Feb. 12 Phy107 Spr07 Lecture 1010Chap. 2: momentum• Momentum = mass × velocity• Momentum can be negative.– For objects moving in opposite directions, one will have positivemomentum and one will have negative momentum– The total momentum could be zero,even though there is plenty of ‘motion’.• Amount of ‘motion’ in a body (but not always positive).• Conservation of momentum:Momentum is not created or destroyed, only transferred fromone object to another.Mon. Feb. 12 Phy107 Spr07 Lecture 1011Conservation of momentum• Useful in understanding result of collisions.• Not concerned with details of collision, only before and after.• Total amount of momentum before= total momentum after.• Momentum is transferredfrom one object to another.Mon. Feb. 12 Phy107 Spr07 Lecture 1012Momentum QuestionA 5 kg ball moving at 1 m/s to the rightcollides with a stationary 10 kg ball.After the collision, the 10 kg ball moves to the right at 0.25 m/s.What is the final speed of the 5 kg ball?A. 0 m/s.B. 0.5 m/sC. 1 m/sD. 0.25 m/sMomentum before= 5kg×1m/s + 10kg×0m/s =5 kg-m/sMomentum after == 5kg× ?m/s + 10kg×0.25m/s =5 kg-m/s= 5kg× ?m/s + 2.5 kg-m/s = 5 kg-m/s? = 0.5 m/s moving to the right3Mon. Feb. 12 Phy107 Spr07 Lecture 1013Momentum exampleTotal momentum before:5 kg × (1 m/s) + 5 kg × (-1 m/s) = 0 kg-m/sAfter collision: total momentum also zero.Balls are stuck together, -> velocity of each must be zero.5 m/s-5 m/s0 m/sBalls stucktogetherbeforeafterMon. Feb. 12 Phy107 Spr07 Lecture 1014Chapter 3• Principle of inertia:– object continues in its state of rest, or uniform motion in astraight line, unless acted upon by a force.• Defined mass m:– amount of inertia of a body– Measured in kg• Define force F:– Something that changes a body’s acceleration• Related force, mass, and acceleration:– F=ma, or a=F/m– Subject to the same force,more massive objects accelerate more slowly.• Weight:– Force of gravity on a body = mg– Measured in newtons (N). 1 N = 1 kg-m/s2Mon. Feb. 12 Phy107 Spr07 Lecture 1015Newton: forces• Newton changed the emphasis from‘before and after’ to ‘during’.• To describe the interaction, he clearlydefined forces and their affects:A force changes the momentum of an object:Change in momentum = Force × timeMon. Feb. 12 Phy107 Spr07 Lecture 1016Question: forceA car weighs 10,000 N.It is moving at a speed of 30 m/s.You apply the brakes with a force of 500 N.How many seconds will it take to stop?A. 10 secondsB. 20 secondsC. 30 secondsD. 60 secondsThe force is 100N.The mass = weight/g = 10,000/10 m/s2 = 1000 kg.So the acceleration is 500 N /1000!kg = 0.5 m/s/s.It takes 60 seconds for the speed to drop from 30m/s to zero.Mon. Feb. 12 Phy107 Spr07 Lecture 1017Newton: 2nd law• Second law is equivalent toNet Force = Mass x Acceleration ! Acceleration =Net ForceMass! a =Fm•Constant force gives constant accel:Velocity increases with time.Mon. Feb. 12 Phy107 Spr07 Lecture 1018Newton: 3rd Law- Action/Reaction• Every force is an interaction between two objects• Each of the bodies exerts a force on the other.• The forces are equal and opposite– An action-reaction pair.Force on theblock by youForce by theblock on youand the earth!4Mon. Feb. 12 Phy107 Spr07 Lecture 1019Newton & mom. conservation• The law of force pairs is the same asconservation of momentum.• An applied force changes the momentum ofan object.• That momentum was transferred from theobject applying the force.• Hence an equal and opposite force had tochange the momentum of the
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