Yung Tae Kim (Physics PhD) Physicist and Educator Puget Sound Community School Seattle, WA

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1 PHYSICIST PROFILE Yung Tae Kim (Physics PhD) Physicist and Educator Puget Sound Community School Seattle, WA Why Physics? Tae, who enjoyed mathematics in high school, signed up for an honors physics class in college. The class only had 8 students and a great professor. [The physics class] was personalized, and it had a huge impact on me. Tae became a physics professor, but wanted to keep that personalized dynamic alive for his students. He turned his advanced classes into workshops. Eventually, Tae became a consultant in video games, providing his skateboarding physics expertise to titles such as Tony Hawk: SHRED and RIDE. Clayton Hauck Now, he serves on the advisory board for a community school geared toward learning at the student s pace. I don t have a career I work on things that I find interesting, and my training in science has given me that adaptability.

2 PHYSICIST PROFILE

3 PHYS 1301 Study Groups z.umn.edu/csewisephysics Connect with other students in your physics lecture You choose when and where to meet Attend a kick off with tutors! Tuesday, October 10, 6-7:30 p.m. 101 Walter Library Supported by the CSE WISE Initiative, csewise@umn.edu

4 A new student group aimed to bridge the gap between Carlson and CBS by hosting speakers and conducting workshops with speakers from science industries throughout the semester Our first speaker, Gina Gettelman, is a program manager at Medtronic When: Tuesday, October 10 from 5:30-6:30PM Where: Coffman 304 Free Jimmy Johns Sandwiches at the meeting!

5 Class 15 October 4, 2017 Forces

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11 Reprise: Newton s Second Law Force is the time derivative of momentum. Momentum is the product of two factors, mass and velocity. There are two possibilities: I: Mass is constant and velocity is variable II: Both mass and velocity are variable (rocket) Case I results in the usual statement of Newton s Second Law F = ma Case I : F = dp dt = d(mv) = m dv dt dt = ma Case II : F = dp dt = d(mv) dt = m dv dt + v dm dt Rocket : 0 = ma + v dm dt 11

12 Gravity Near the surface of the Earth, the acceleration of a mass due to gravity is g = 9.81 m-s -2 towards the center of the Earth. Since F = ma and a = g, then the force of gravity is F gravity = mg with F pointing towards the center of the Earth. The gravitational force can be represented in the context of components, which is particularly useful for problems with inclined planes. For an inclined plane making an angle q with the horizontal, the components of gravity are mg sin q, along the plane, mg cos q, perpendicular (normal) to the plane. q 12

13 Problem A 2.00 kg mass with an initial velocity of 2.00 m/s slides down a track inclined at 5 above the horizontal. Because of friction, the mass comes to a stop after sliding for 4.00 s. What is the total impulse? (a) kg-m/s (b) kg-m/s (c) kg-m/s (d) kg-m/s (e) kg-m/s 13

14 Problem A 2.00 kg mass with an initial velocity of 2.00 m/s slides down a track inclined at 5 above the horizontal. Because of friction, the mass comes to a stop after sliding for 4.00 s. What is the total impulse? (d) kg-m/s The total impulse is the change in momentum. The initial momentum is mv = (2 kg)(2 m/s) = 4 kg-m/s. The final momentum is zero. The change is kg-m/s. This total impulse results from two forces Gravity accelerates the forward motion Friction retards the forward motion Since the mass stops, for this problem, friction is stronger than gravity. 14

15 Problem A 2.00 kg mass with an initial velocity of 2.00 m/s slides down a track inclined at 5 above the horizontal. Because of friction, the mass comes to a stop after sliding for 4.00 s. What is the average total force? (a) N (b) N (c) N (d) N (e) N 15

16 Problem A 2.00 kg mass with an initial velocity of 2.00 m/s slides down a track inclined at 5 above the horizontal. Because of friction, the mass comes to a stop after sliding for 4.00 s. What is the total force? (a) N The force in this problem appears to have a constant magnitude. Then, the equation for impulse becomes Dp = FDt So, (-4.00 kg-m/s) = F(4.00 s), F = N 16

17 Problem A 2.00 kg mass with an initial velocity of 2.00 m/s slides down a track inclined at 5 above the horizontal. Because of friction, the mass comes to a stop after sliding for 4.00 s. How far down the track does the mass travel before it comes to a stop? (a) 1.00 m (b) 2.00 m (c) 3.00 m (d) 4.00 m (e) 5.00 m 17

18 Problem A 2.00 kg mass with an initial velocity of 2.00 m/s slides down a track inclined at 5 above the horizontal. Because of friction, the mass comes to a stop after sliding for 4.00 s. How far down the track does the mass travel before it comes to a stop? (d) 4.00 m d =(2.00)(4.00) F = ma d = v 0 t at2 = v 0 t a = F m F m t (4.00)2 =4.00 m 18

19 Problem A 2.00 kg mass with an initial velocity of 2.00 m/s slides down a track inclined at 5 above the horizontal. Because of friction, the mass comes to a stop after sliding for 4.00 s. What is the force of friction? (a) 2.71 N (b) 3.04 N (c) 3.25 N (d) 3.89 N (e) 4.11 N 19

20 Problem A 2.00 kg mass with an initial velocity of 2.00 m/s slides down a track inclined at 5 above the horizontal. Because of friction, the mass comes to a stop after sliding for 4.00 s. What is the force of friction? (a) 2.71 N F total = F gravity + F friction F gravity = ma = mg sin =(2.00 kg)(9.81 m/s 2 )sin5 F gravity = 1.71 N F friction =1.00 N N =2.71 N 20

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