13) Two skaters, an 82-kg man and a 48-kg woman, are standing on ice. The woman pushes on the man with a force of 45 Newtons due east.

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1 Lab 14: Two Ball Race ; Egg Drop - The Law of Force and Acceleration (Newtonʼs 2nd Law) - Distance and velocity vs. time graphs for constant acceleration - Solving problems with force, mass and acceleration 13) Two skaters, an 82-kg man and a 48-kg woman, are standing on ice. The woman pushes on the man with a force of 45 Newtons due east. a) What are the accelerations of the man and the woman? There are equal and opposite forces on both the man and woman due to the Law of Interaction: force = mass acceleration acceleration = force / mass for the man: acceleration = 45 N / 82 kg = 0.54 m/s 2 for the woman: acceleration = 45 N / 48 kg = 0.94 m/s 2 b) In which directions will the man and woman accelerate? They will push apart and accelerate in opposite direction - the man will accelerate east, the woman will accelerate west. c) How would this situation change, if at all, if the man pushed on the woman with a force of 45 Newtons due west? There would be no change at all - the Law of Interaction dictates that there will be a 45 Newton force in equal and opposite directions, in either case. 14) Joseph drops a nickel, and observes that it accelerates downward. He points out to his lab partners that since it is accelerating, the forces on the nickel are not balanced. Chantel agrees with him on this point. Then Joseph says the Law of Interaction doesnʼt apply to the nickel - that there is no equal and opposite force, because it is accelerating. Chantel disagrees - she insists if gravity is pulling the nickel down to the earth, the nickel must be pulling the earth upward with an equal and opposite force. Joseph says she is wrong, because we donʼt notice the earth accelerating upward.

2 a) If the nickel weighs 0.05 Newtons, and Chantel is correct, with what force is the nickel pulling on the earth? If Chantel is correct, the nickel pulls the earth upward with a 0.05 N force, just as the earth pulls the nickel downward with a 0.05 N force. b) Joseph looks up the mass of the earth on the internet, and finds it is about 6,000,000,000,000,000,000,000,000 kg (thatʼs 6 million billion billion kg!!!!) Determine (or simply estimate) the acceleration of the earth due to the force in part a), using the Law of Force and Acceleration. force = mass acceleration acceleration = force / mass = 0.05 N / 6,000,000,000,000,000,000,000,000 kg = m/sec 2 If you donʼt feel like calculating this exactly like I did, you can estimate that the value is much less than a millionth of a billionth of a billionth of a m/sec 2 just given the numbers - it suffices to say that this is an IMMEASURABLY small acceleration! c) Given your answer to part b), is Josephʼs argument valid? Who is correct? Joseph is wrong - the equal and opposite force of 0.05 N on the earth would not produce any acceleration they would notice because of the earthʼs HUGE mass; Chantel is correct. d) After dropping the nickel, Joseph picks it up again, exerting a force of 0.05 Newtons upwards to lift it up. Is there an equal and opposite force on the earth in this case, according to the Law of Interaction? Explain. Yes - as Joseph pulls the nickel up again with 0.05 N, he must exert an equal and opposite force back down on the earth, in the form of an extra (very small) push of 0.05 N through the bottom of his feet.

3 Lab 15: Force and Energy ; Drop the Ball - Calculating energy as force distance - Kinetic energy - Conservation of energy 15) Joseph (from Question 13) lifts the 0.05 Newton nickel to a height of 2 meters before dropping it. a) How much energy is stored in the nickel before he drops it? What is the nickelʼs kinetic energy at this point? energy = force distance = 0.05 N 2 m = 0.1 Joules b) After the nickel falls 1 meter, how much stored energy does it have left? energy lost = force distance fallen = 0.05 N 1 m = 0.05 Joules; 0.1 Joules 0.05 Joules = 0.05 Joules left c) What is the kinetic energy of the nickel after it falls 1 meter? The missing energy was turned into kinetic energy, so the kinetic energy is 0.05 Joules. d) What is the speed of the nickel after it falls 1 meter? kinetic energy = ½mass velocity if we do some algebra: velocity 2 = 2 kinetic energy / mass velocity = (2 kinetic energy / mass) [ is a square root sign ] = ( Joules / kg) = (20) m/sec = 4.47 m/sec

4 e) Can you use stored and kinetic energy to find the speed of an nickel dropped from ANY height? Explain. (Use an example if you prefer.) Yes - you just need to know how much of the initial energy was lost, which you can figure out from the height. Then, you use the kinetic energy formula to determine its velocity as in part d. For example, if the nickel fell 1.5 meters, it would only have Joules of stored energy left; the kinetic energy would then be Joules - the rest of the calculation would be the same. f) In part e), does the mass of the object matter? Why or why not? Although mass appears in the energy formulas, it turns out the mass of a following object does not contribute to speed, because energy = force distance = mass acceleration distance kinetic energy = ½mass velocity 2 Equating these gives mass acceleration distance = ½mass velocity 2 canceling mass on both sides gives acceleration distance = ½velocity 2 You can solve for velocity in this; the mass of a falling object doesnʼt matter (unless high air resistance is involved - this is another problem entirely...) 16) In this lab activity, you were given an example of a strong man trying to stop a moving freight car by pulling on a rope. If the freight car has a mass of kg and is moving at 4 meters per second, and the man can pull with a force of 1000 Newtons, and the end of the rails is 100 meters away, will he stop the train in time? If not, how much extra track would he need?

5 First, find the kinetic energy of freight car: kinetic energy = ½mass velocity 2 = ½ kg 4 m/s 4 m/s = Joules This is the amount of energy needed to stop the train: energy = force distance distance = energy / force = Joules / 1000 N = 160 m So, the man needs 160 meters of track to stop the freight car, but only has 100 meters of track; he would need an extra 60 m of track. Lab 16: Some more examples of force, motion, and energy ; Friction - Force and distance in pulleys - Energy changing forms - Force of friction 17) Two pulley systems are shown in the above diagrams, both connected to a 1000 Newton weight.

6 a) How much energy is required to lift a 1000 Newton weight a height of 1 meter? energy = force distance = 1000 N 1 m = 1000 Joules b) The left pulley has one length rope for the pull (on the far left, marked by an arrow), and two lengths of rope for support. If the weight is lifted, how much force will be on either of the pull ropes in this system? (Draw this on the diagram.) 1000 N / 2 = 500 N c) Since you only need to pull one rope to lift the weight, how much force will one need to pull with to left the weight in the left pulley system? 500 N d) Knowing that the energy = force distance is constant for lifting a weight, using your answers from a) and c), what length of rope will one need to pull to lift the weight a height of 1 meter in the left pulley system? energy = force distance distance = energy / force = 1000 Joules / 500 N = 2 m e) Using the same procedure you used in a)-d), determine the force and length of rope required to lift the weight in the right pulley system to a height of 1 meter N / 3 = 333 N distance = energy / force = 1000 Joules / 333 N = 3 m

7 18) A 5.00 kg bowling ball, starting at rest, rolls down a ramp from a height of 2.00 meters. When it reaches the ground, it is rolling at a speed of 1.95 m/sec. How much energy in the bowling ball was lost to the surroundings as a result of friction? weight = mass 10 m/s 2 = 5.00 kg 10 m/s 2 = 50.0 N initial energy = force distance = 50.0 N 2.00 m = Joules At the end, all the remaining energy is kinetic energy: final energy = ½mass velocity 2 = kg 1.95 m/s 1.95 m/s = 9.5 Joules This means that = 90.5 Joules was lost as friction. This is not a realistic number! This is because thereʼs a typo in the original version - using a velocity of 6 m/s gives a more realistic answer: final energy = ½mass velocity 2 = kg 6 m/s 6 m/s = 90 Joules so that only = 10 Joules was lost to friction.