2. Physics 101: Lecture 4, Pg 1 Kinematics + Dynamics Physics 101: Lecture 04 l Today’s lecture will cover Textbook Chapter 4 Exam I

3. Physics 101: Lecture 4, Pg 2 What’s Most Difficult I’m a little slow dusting off the trig skills. Everything :( …I think I just need some more practice walking through the thought process required to solve the problems. I LOVE Free Body Diagrams. The most difficult problems right now are 2 dimension analysis problems where you don’t know 2 variables so you have to end up substituting in one variable for the other and manipulating equations. I am having a hard time with vectors in this class… 05 Identifying all forces involved with free body diagrams. The lectures don’t seem to explain much, just random letters and numbers…

4. Physics 101: Lecture 4, Pg 3 Review l Kinematics : Description of Motion è Position è Displacement  Velocity v =  x /  t »average »instantaneous  Acceleration a =  v /  t »average »Instantaneous 07

5. Physics 101: Lecture 4, Pg 4 Preflight 4.1 (A) (B)(C) Which x vs t plot shows positive acceleration? 08 90% got this correct!!!! “The slope of the graph is getting increasingly positive; the velocity is increasing over time.” …most trouble with deciphering graphs…

6. Physics 101: Lecture 4, Pg 5 Equations for Constant Acceleration (text, page 108) l x = x 0 + v 0 t + 1/2 at 2 l v = v 0 + at l v 2 = v 0 2 + 2a(x-x 0 ) l  x = v 0 t + 1/2 at 2 l  v = at l v 2 = v 0 2 + 2a  x Lets derive this… 10

7. Physics 101: Lecture 4, Pg 6 Equations for Constant Acceleration (text, page 108) l x = x 0 + v 0 t + 1/2 at 2 l v = v 0 + at l v 2 = v 0 2 + 2a(x-x 0 ) l  x = v 0 t + 1/2 at 2 l  v = at l v 2 = v 0 2 + 2a  x 14

8. Physics 101: Lecture 4, Pg 7 Equations for Constant Acceleration (text, page 80) l x = x 0 + v 0 t + 1/2 at 2 l v = v 0 + at l v 2 = v 0 2 + 2a(x-x 0 ) l  x = v 0 t + 1/2 at 2 l  v = at l v 2 = v 0 2 + 2a  x 14

9. Physics 101: Lecture 4, Pg 8 Kinematics Example l A car is traveling 30 m/s and applies its breaks to stop after a distance of 150 m. How fast is the car going after it has traveled ½ the distance (75 meters) ? A) v 15 m/s 18 This tells us v 2 proportional to  x

10. Physics 101: Lecture 4, Pg 9 Kinematics Example l A car is traveling 30 m/s and applies its breaks to stop after a distance of 150 m. How fast is the car going after it has traveled ½ the distance (75 meters) ? A) v 15 m/s 18

11. Physics 101: Lecture 4, Pg 10 Acceleration ACT An car accelerates uniformly from rest. If it travels a distance D in time t then how far will it travel in a time 2t ? A. D/4 B. D/2 C. D D. 2D E. 4D Correct x=1/2 at 2 Follow up question: If the car has speed v at time t then what is the speed at time 2t ? A. v/4 B. v/2 C. v D. 2v E. 4v Correct v=at Demo… 24

12. Physics 101: Lecture 4, Pg 11 Chocolate l Advantages è Tastes Good! è Health Benefits (similar to red wine) è Releases good chemicals in brain è Improves Learning!!!!!!! l Disadvantages è Cost è Distribution 26

13. Physics 101: Lecture 4, Pg 12 Newton’s Second Law  F=ma

14. Physics 101: Lecture 4, Pg 13 ACT l A force F acting on a mass m 1 results in an acceleration a 1.The same force acting on a different mass m 2 results in an acceleration a 2 = 2a 1. What is the mass m 2 ? (A) (B) (C) (A) 2m 1 (B) m 1 (C) 1/2 m 1 Fa1a1 m1m1 F a 2 = 2a 1 m2m2 F=ma F= m 1 a 1 = m 2 a 2 = m 2 (2a 1 ) Therefore, m 2 = m 1 /2 Or in words…twice the acceleration means half the mass 29

15. Physics 101: Lecture 4, Pg 14Example: y x 35 X direction: Tractor  F = ma F w – T = m tractor a F w = T+ m tractor a T W N X direction: Trailer  F = ma T = m trailer a T FwFw W N Combine: F w = m trailer a + m tractor a F w = (m trailer+ m tractor ) a A tractor T (m=300Kg) is pulling a trailer M (m=400Kg). It starts from rest and pulls with constant force such that there is a positive acceleration of 1.5 m/s 2. Calculate the horizontal force on the tractor due to the ground. F W = 1050 N

16. Physics 101: Lecture 4, Pg 15 Net Force ACT Compare F tractor the net force on the tractor, with F trailer the net force on the trailer from the previous problem. A) F tractor > F trailor B) F tractor = F trailor C) F tractor < F trailor  F = m a F tractor = m tractor a = (300 kg) (1.5 m/s 2 ) = 450 N F trailer = m trailer a = (400 kg) (1.5 m/s2) = 600 N 38

17. Physics 101: Lecture 4, Pg 16 Pulley Example l Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. 1 2 1) T - m 1 g = m 1 a 1 2) T – m 2 g = -m 2 a 1 2) T = m 2 g -m 2 a 1 1) m 2 g -m 2 a 1 - m 1 g = m 1 a 1 a 1 = (m 2 – m 1 )g / (m 1 +m 2 ) 1 T m1gm1g 2 T m2gm2g Compare the acceleration of boxes 1 and 2 A) |a 1 | > |a 2 | B) |a 1 | = |a 2 | C) |a 1 | < |a 2 | y x 42

18. Physics 101: Lecture 4, Pg 17 Pulley Example l Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. 1 2 a 1 = (m 2 – m 1 )g / (m 1 +m 2 ) a = 2.45 m/s 2  x = v 0 t + ½ a t 2  x = ½ a t 2 t = sqrt(2  x/a) t = 0.81 seconds 1 T m1gm1g 2 T m2gm2g Compare the acceleration of boxes 1 and 2 A) |a 1 | > |a 2 | B) |a 1 | = |a 2 | C) |a 1 | < |a 2 | y x 46

19. Physics 101: Lecture 4, Pg 18 Summary of Concepts Constant Acceleration  x = x 0 + v 0 t + 1/2 at 2  v = v 0 + at  v 2 = v 0 2 + 2a(x-x 0 ) F = m a – Draw Free Body Diagram – Write down equations – Solve I feel that if there were suggested homework problems from the book, I would have a lot more practice with the concepts we’re learning. Ch 2 Problems: 1,5,7,11,13,17,29,49,69 Ch 3 Problems: 5,13,33,47,57,65,67