PHY 102 1st EditionOutline of Last LectureI. Conservation of Momentuma. Experimentb. ExampleII. Numerical Problemsa. Example 1b. Example 2c. Example 3III. Newton’s Second LawIV. Second Law RepriseOutline of Current Lecture I. Examples of Newton’s Second Lawa. Car accelerationb. Collision Examplec. Weightd. Change in VelocityII. Experience and Experimenta. Experience shows that:b. Experiment Shows that:III. Newton’s Third LawThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Current LectureI. Continued examples of Newton’s Second Lawa. A car with the mass of 1500 kg accelerates at a rate of -2 m/s2 under a constant force. What are the magnitude and direction of that force?∑F=maForce = ?M = 1500 kgA = -2 m/s2F = 1500 kg * -2 m/s2 = -3000 N N = kg*m/s2b. How much force would it take to slow a 75 kg person riding in a car going 20 m/s to a complete stop if the time was .03 seconds (the typical time of an auto collision)?∑F=maM = 75 kgV = 20 m/sT = .03 s a = ∆ v∆ tF∆t = m ∆ v or F = m∆v/∆ tvf – vi = 0 m/s – 20 m/s∆ v = - 20m/s75kg∗−20 m/s.03 sc. How much upward force does the ground apply to someone with a mass of 75 kgto counter balance the pull of gravity? Note that g = -9.8 m/s2 and that Fnet = ma and that a = g.F = ma Wt = mg (Weight = mass times 9.81m/s2)F = ?M = 75 kgA = g = -9.8 m/s2(75 kg)(-9.8 m/s2) = 740 N-740 N Weight of the person740 N Force of the earth on the persond. A force of 7 N is applied to a 3.5 kg mass for 2 seconds. What is the change of velocity?F = maF = 7 N F = m * ∆ v∆ t∆ v=?M = 3.5 kg∆ v=(7.0 N )(2 s)3.5 kgT = 2 s 4 m/sII. Experience & Experimenta. Experience shows thati. A greater force for a given mass causes greater acceleration – same force applied to two masses gives different resultsii. A greater mass for a given force gives lesser accelerationb. Experiment shows that:i. Acceleration is directly proportional to forcec.∑F=mai. A newton is the force required to accelerate a 1 kg mass at a rate of 1 m/s2.III. Newton’s Third Lawa. Specific case of the second lawi.∑F=ma – the first law can be derived from this; just means the acceleration is 0b. F12 = -F21i. Most misunderstood of the 3 lawsii. If an object A exerts a force Fab on an object B, then object B exerts a forceFba on object A such that Fab = -Fbaiii. The sum of the two forces (vectors) have to be 0 because the object is notaccelerating.Large mass small massThe change in velocity will be less in the large truck because it has a larger mass so the passenger will likely survive.Forces are equal in the event of a collisioniv. Tug of war is possible because you have to consider all the forces on the rope, not just the tension on the rope. If the only force was the tension on the rope, no one would ever win.Centripetal force = center-seeking
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