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Kinematics + Dynamics Today’s lecture will cover Textbook Chapter 4 Physics 101: Lecture 04 Exam I

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Review Kinematics : Description of Motion – Position – Displacement – Velocity v = x / t average instantaneous – Acceleration a = v / t average Instantaneous 07

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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…

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Equations for Constant Acceleration (text, page 108) x = x 0 + v 0 t + 1/2 at 2 v = v 0 + at 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

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Equations for Constant Acceleration (text, page 108) x = x 0 + v 0 t + 1/2 at 2 v = v 0 + at 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

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Equations for Constant Acceleration (text, page 80) x = x 0 + v 0 t + 1/2 at 2 v = v 0 + at 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

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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 18 This tells us v 2 proportional to x

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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 18

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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

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Chocolate Advantages – Tastes Good! – Health Benefits (similar to red wine) – Releases good chemicals in brain – Improves Learning!!!!!!! Disadvantages – Cost – Distribution 26

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Newton’s Second Law F=ma

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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

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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 2. Calculate the horizontal force on the tractor due to the ground. 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 F W = 1050 N

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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

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Pulley Example 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

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Pulley Example 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

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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

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Physics 101: Lecture 4, Pg 1 Kinematics + Dynamics Physics 101: Lecture 04 l Today’s lecture will cover Textbook Chapter 4 iclicker scores have been imported.

Physics 101: Lecture 4, Pg 1 Kinematics + Dynamics Physics 101: Lecture 04 l Today’s lecture will cover Textbook Chapter 4 iclicker scores have been imported.

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