Kinematics + Dynamics

Transcription

1 Exam I Physics 101: Lecture 04 Kinematics + Dynamics Today s lecture will cover Textbook Chapter 4 iclicker scores have been imported again please check! If you are new to the course, please read the course description on the course web page (and policy from Lecture 1 note)! neptune Physics 101: Lecture 4, Pg 1

2 What s s Most Difficult I am finding vectors as the most difficult problems in class right now. The concept with the kinetic and static forces. Trying to remember the formulas and im rusty with trig so all the sin, cos, and tan is tricky for me. Tension / setting up problem / vectors / free body diagrams. pulley problems remembering to do the prefilghts / Staying awake. Physics 101: Lecture 4, Pg

3 You can review vectors using the textbook!!! Pages 8-35 of Textbook This also includes trigonometry Also see course web site. Physics 101: Lecture 4, Pg 3

4 Discussion book answers are posted online before the session! Quiz answers posted too but after the session. Physics 101: Lecture 4, Pg 4

5 Review Kinematics : Description of Motion Position Displacement Velocity v = x / t» average» instantaneous Acceleration a = v / t» average» instantaneous Relative velocity: v ac = v ab + v bc Physics 101: Lecture 4, Pg 5

6 Preflight 4.1 interpreting graphs (A) (B) (C) Which x vs t plot shows positive acceleration? 89% got this correct!!!! This shows that more distance is being covered per second as the graph proceeds. This means that the speed of the car is increasing which means a positive acceleration. Physics 101: Lecture 4, Pg 6

7 Equations for Constant Acceleration (text, page ) 114) x (meters) x = x 0 + v 0 t + 1/ at x = v 0 t + 1/ at t (seconds) v (m/s) 0 v = v 0 + at v = v 0 + a(x-x 0 ) x x 0 v t vv0 v vf f v0 ( v0 v v 0 at 1 v0 at 0 v v = at v = v 0 + a x x ) x 0 v 0 t 1 at t (seconds) a (m/s ) t (seconds) Physics 101: Lecture 4, Pg 7 14

8 Kinematics Example 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 B) v = 15 m/s C) v > 15 m/s a v vo a x v f v o (150) 30 (150) v 30 a (75) 75 ( 30 ) v75 30 (75) (150) 1 v75 30 ( 30 ) 1 v75 30 v m / s Physics 101: Lecture 4, Pg 8

9 Acceleration ACT A car accelerates uniformly from rest. If it travels a distance D in time t then how far will it travel in a time t? A. D/4 B. D/ C. D D. D Demo E. 4D Correct x=1/ at Follow up question: If the car has speed v at time t then what is the speed at time t? A. v/4 B. v/ C. v D. v Correct v=at E. 4v Physics 101: Lecture 4, Pg 9

10 Newton s s Second Law F=ma position and velocity depend on history Net Force determines acceleration Physics 101: Lecture 4, Pg 10

11 ACT A force F acting on a mass m 1 results in an acceleration a 1.The same force acting on a different mass m results in an acceleration a = a 1. What is the mass m? m 1 m F a 1 F a = a 1 (A) m 1 (B) m 1 (C) 1/ m 1 F=ma F= m 1 a 1 = m a = m (a 1 ) Therefore, m = m 1 / Or in words twice the acceleration means half the mass Physics 101: Lecture 4, Pg 11

12 Example: 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. Calculate the horizontal thrust force on the tractor due to the ground. X direction: Tractor F = ma F Th T = m tractor a F Th = T+ m tractor a X direction: Trailer F = ma T = m trailer a N F Th = 1050 N T T W W Combine: F Th = m trailer a + m tractor a F Th = (m trailer+ m tractor ) a Physics 101: Lecture 4, Pg 1 y N x F Th

13 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 F = m a B) F tractor = F trailor F tractor = m tractor a C) F tractor < F = (300 kg) (1.5 m/s ) trailor = 450 N F trailer = m trailer a = (400 kg) (1.5 m/s) = 600 N Physics 101: Lecture 4, Pg 13

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15 Pulley Example Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box has a mass of.5 kg. Box starts from rest 0.8 meters above the table, how long does it take to hit the table. y Compare the acceleration of boxes 1 and x A) a 1 > a B) a 1 = a C) a 1 < a a 1 = (m m 1 )g / (m 1 +m ) a =.45 m/s x = v 0 t + ½ a t x = ½ a t t = sqrt( x/a) t = 0.81 seconds 1 T m 1 g T m g 1 Physics 101: Lecture 4, Pg 15

16 Summary of Concepts Constant Acceleration x = x 0 + v 0 t + 1/ at v = v 0 + at v = v 0 + a(x-x 0 ) F = m a Draw Free Body Diagram Write down equations Solve Next time: textbook section 4.3, 4.5 Physics 101: Lecture 4, Pg 16