1Welcome to Physics 1010: The Physics of Everyday Life www.colorado.edu/pjysics/phys1010You are responsible for bringing a working clicker every day.HW 3 is available on CULearnNext week: class will be covered by Rachel PepperToday• Skydiving (Net force)• Crashing Cars (Ouch)• Driving in snow (Friction)Confusing language Regular people: acceleration: speeding up.Physicists: acceleration: the rate of speeding up or slowing down or changing direction of motion.Regular people: kg: a weight= 2.2 pounds,Physicists: kg: a mass. On earth this mass has a weight of 1kg x 9.8 m/s2 = 9.8 NRegular people: Velocity: how fast you are going = speed.Physicists: Velocity: the speed and direction of motion.We have a bunch of unregistered clickers. Register clicker today! Terminal velocityA 100 kg man jumps from a plane. Immediately after he jumps,a. The force of gravity on the man is 100 kg downwards, the net force on him is 100kg downwards, and his acceleration is 9.8 m/s2downwards.b. The force of gravity on the man is 100 kg downwards, his net force is 0, and his acceleration is 0 m/s2. c. The force of gravity on the man is 980 Newtons downwards, the net force on him is 980 Newtons downwards, and his acceleration is 9.8 m/s2downwards.d. The force of gravity on the man is 980 Newtons, the net force on the man is 0 Newtons, and his velocity increases until it is 9.8 m/s downwards and then stops increasing. e. The force of gravity on the man is 980 Newtons, the net force on the man is 0 Newtons, and his acceleration is 9.8 m/s2downwards. A 100 kg man jumps from a plane. Immediately after he jumps,c. The force of gravity on the man is 980 Newtons downwards, the net force on him is 980 Newtons downwards, and his acceleration is 9.8 m/s2downwards.Forces always measured in Newtons! 100 kg is a mass. Force of gravity on an object = (mass of object) x (9.8 m/s2) and is straightdownwards. = 100 kg x 9.8 m/s2= 980 Newtons Sum of all forces = Fnet  In this case, Fnet= Fgravity on man Fnet= mass x accelerationSince Fnetis not zero, tells us man will accelerate! Acceleration is always in the same direction as the net force! Fgravity on man2As man falls for some time, he finds that he is falling at a constant velocity. At this time… a. The force of gravity on the man is 100 kg downwards, the net force on him is 100kg downwards, and his acceleration is 9.8 m/s2downwards.b. The force of gravity on the man is 100 kg downwards, his net force is 0, and his acceleration is 0 m/s2. c. The force of gravity on the man is 980 Newtons downwards, the net force on him is 980 Newtons downwards, and his acceleration is 9.8 m/s2downwards.d. The force of gravity on the man is 0 Newtons, the net force on the man is 0 Newtons, and his acceleration is 0 m/s2. e. The force of gravity on the man is 980 Newtons, the net force on the man is 0 Newtons, and his acceleration is 0 m/s2. As man falls for some time, he finds that he is falling at a constant velocity. At this time… e. The force of gravity on the man is 980 Newtons, the net force on the man is 0 Newtons, and his acceleration is 0 m/s2. Force of gravity = (mass of object) x (9.8 m/s2) on an object (on Earth) = 100 kg x 9.8 m/s2= 980 Newtons and is straightdownwards. Sum of all forces = Fnet  In this case, Fnet= Fgravity on man + Fair on man = 0 !!!!Fnet= mass x accelerationSince Fnetis zero, tells us man will not accelerate! Acceleration will be zero, Velocity will be constant ! Fgravity on manFair on manSum of all forces = Fnet  In this case, Fnet= Fgravity on man + Fair on manFnet= mass x accelerationSince Fnetis not zero, tells us man will accelerate! Acceleration will be smaller than 9.8 m/s2because net force is smaller than force of gravity on man. Velocity will be changing ! Speeding up in downwards directionFgravity on manFair on manBefore reaching terminal velocity, forces don’t quite cancel. Still small acceleration.FnetCrashing little cars: Case A: hit hard end of rampSPONGECase B: hit sponge at end of rampSketch your predictions of the velocity vs. time, acceleration vs time, net force, and measured force (by probe) for this motion. Starting when we let go and ending after crash. FOR CASE A AND B on same graph. VelocitytimeNetForcetimeMeasure ForcetimeAccelerationtimeCrashing little carsVelocitytimeNetforcetimeMeas ForcetimeAccelerationtimeVelocitytimeNetforcetimeMeasForcetimeAccelerationtimeD. Same as a, but with all curves on negative side of 0 instead of positive.E. None of these choices. BCCase A: hit hard end of rampSPONGECase B: hit sponge at end of rampVelocitytimeNetForcetimeMeas. ForcetimeAccelerationtimeACrashing little cars: Case 1: hit hard end of rampSPONGECase 2: hit sponge at end of rampAccelerationduring crash= change in velocitytime elapsed during crashJust before crash: Car in case 1 is at same velocity as Car in case 2After crash:Both car in 1 and 2 are at rest. Velocity = 0 m/sShorter in Case 1Than Case 2Same in bothCase 1 and 2So, during crashacceleration of 1 greater than acceleration of 2. Forcein 1 greater … ouch!VelocitytimeNetForcetimeMeas. ForcetimeAccelerationtimeAWall exerts force on car.3Force sensor plus metal weight has massof 1.2 kg, It weighs approx. how many N? (How much force needed to lift it?)a. 1.2 b. 12 c. 1.2/2.2 d. (1.2/2.2) x 9.8PULL Force sensor + metal weightFriction between table and force sensor. Predict graph of force which we must apply by pulling on string in order to move sensor along table at a constant speed … prediction should include force from before starting to pull until sensor is moving at constant speed across the table. (Make guess as to specific value as well as shape.)AppliedForcetimePositive direction? N-? NAnswer: bPULL Force sensor + metal weightFriction between table and force sensor. Predict graph of force which we must apply by pulling on string in order to move sensor along table at a constant speed … prediction should include force from before starting to pull until sensor is moving at constant speed across the table. timeAppliedForcetimeAppliedForcetimeAppliedForceStarts movinghereABCPULL Force sensor + metal weighttimeForcesStarts movinghereAApplied ForceFriction ForceStatic frictionSliding frictionBalancing sliding friction: How much force is required to keep it moving along table at constant speed if weight of (force sensor + metal ) = mg = 12 N ?a. Weight x 0.3, b. Weight x 0.7, c.