NEWTON’S LAWS OF MOTION MIT C.P. PHYSICS NEWTON’S LAWS OF MOTION

Law of Universal Gravity Sir Isaac Newton Law of Universal Gravity

Lab: Hey Newton, Pass the Water!

Lab: Hey Newton, Pass the Water! Data Chart Volume (mL) Finish Order

Newton’s Laws of Motion First Law of Motion An object at rest will stay at rest unless acted upon by an outside force. An object in motion will stay in motion unless acted upon by an outside force.

Newton’s 1st Law of Motion Inertia

Inertia increases with mass.

Lab: Inertia-It Makes Cents!

Lab: Inertia Ball

Lab: Inertia Ball

Lab: Inertia Ball Data Chart Pull Slowly Quickly

Lab: Inertia Ball Analysis Pull Slowly Quickly

Applications

Applications

Lab: Newton’s 2nd Law of Motion Interactive

Lab: Newton’s 2nd Law of Motion Data Chart Trial # Mass of Wagon (g) Hanging Mass Acceleration (m/s/s) 1 100 2 3 4 5 200 6 400 7 600 8 800 9 1000

Lab: Newton’s 2nd Law of Motion Calculation Chart Trial # Mass (kg) F (N) Acceleration (m/s/s) 1 2 3 4 5 6 7 8 9

Lab: Newton’s 2nd Law of Motion Formulas Mass (kg) = mass of wagon/1000 Force (N) = hanging mass X 0.0098

Newton’s Laws of Motion 2nd Law of Motion The sum of all forces acting on an object, F(net), equals the object’s mass times its acceleration. F(net) = ma

Newton’s 2nd Law of Motion Small F(net), small m = acceleration Small F(net), big m = no acceleration

Sample Problem 1 Two students are sitting across each other in lunch. Student A slides a 2.2kg plate of food towards student B. If the net force is 1.6N to the right, what will be the plate’s acceleration? Step 1 F(net) = ma Step 2 1.6N = (2.2 kg)(a) Step 3 a = 0.73 m/s/s

Sample Problem 2 The net force acting on a model airplane is 7.0N. The plane accelerates at a rate of 2.2 m/s/s. What is the mass of the model airplane? Step 1 F(net) = ma Step 2 7.0N = (m)(2.2m/s/s) Step 3 m = 3.2 kg

Lab: Football Physics

Lab: Football Physics Data Chart Player Weight (N) 36 m Speed (m/s) 36 m Time (s) Nate 833 16.0 4.51 “T.D.” 735 16.4 4.40 Bubba 911 15.0 4.82 Tony 825 15.3 4.71 Tiny 1010 14.7 4.90 Mike 931 15.7 4.60

Lab: Football Physics Calculation Chart Player Mass (kg) Acceleration (m/s/s) F(net) (N) Nate “T.D.” Bubba Tony Tiny Mike

Lab: Football Physics Analysis Linemen 1. 2. Backs Ends

Lab: The Elevator Lab

If all the forces acting on an object are balanced, the object moves at a constant velocity.

Lab: The Elevator Lab Data Chart Procedure # Force (N) 1 2 3 4 5

 = Balanced  = Unbalanced Lab: The Elevator Lab Calculation Table F(g) (N) Mass (kg) a (m/s/s) F(app)   1 2 3 4 5  = Balanced  = Unbalanced

Sample Problem 1 A student has a mass of 200.0 kg. He wants to lose weight so he stands on a bathroom scale while in an elevator. He presses the up button causing him to ascend at an acceleration of 4.00 m/s/s. What does the bathroom scale read on the way up? Step 1 Determine the F(g). F(g) = mg F(g) = (200. kg)(9.81m/s/s) F(g) = 1960 N

Sample Problem 1 Step 2 Determine the F(app). F(app) = ma F(app) = (200. kg)(4.00 m/s/s) F(app) = 800.N Step 3 Determine the F(net). F(net) = F(g) + F(app) F(net) = 1960N + 800.N F(net) = 2760N

Sample Problem 2 A student has a mass of 200.0 kg. He wants to lose weight so he stands on a bathroom scale while in an elevator. He presses the down button causing him to descend at an acceleration of 4.00 m/s/s. What does the bathroom scale read on the way down? Step 1 Determine the F(g). F(g) = mg F(g) = (200. kg)(9.81m/s/s) F(g) = 1960 N

Sample Problem 2 Step 2 Determine the F(app). F(app) = ma F(app) = (200. kg)(-4.00 m/s/s) F(app) = -800.N Step 3 Determine the F(net). F(net) = F(g) + F(app) F(net) = 1960N + (-800.N) F(net) = 1160N

Free Fall & Terminal Velocity

Terminal Velocity

Sample Problem 1 At a point during a jump, a 89.1 kg skydiving dog experience a F(air) of 400.N Calculate the net force on the skydiving dog. Step 1 F(net) = F(g) + F(air) Step 2 F(net) = (89.1kg)(9.81m/s/s) + (-400N) Step 3 F(net) = 474N

Sample Problem 2 At a point during a jump, a 89.1 kg skydiving dog experience a F(air) of 400.N Calculate the skydiving dog’s rate of acceleration. Step 1 F(net) = ma Step 2 474 = (89.1kg)a Step 3 a = 5.32 m/s/s

Feather & Elephant Challenge

Feather & Elephant Challenge

Newton’s Laws of Motion 3rd Law of Motion For every action, there is always an opposite but equal reaction.

Applications

Action-Reaction Pairs

Action-Reaction Pairs

Which Rope Is Being Pulled With More Force? Same!

Identify 6 Pairs of Action-Reaction

Test Review Game Added Points Score(Single) Score(Double) 1 3000 1500 2 3500 1750 3 4000 2000 4 4500 2250 5 5000 2500 6 5500 2750 7 6000 8 6500 3250 9 7000 10 7500 3750