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Newton’s Second Law The Mathematical One. What is the relationship? ForceMassAcceleration Force  Constant  Mass  Constant  Acceleration  Constant.

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Presentation on theme: "Newton’s Second Law The Mathematical One. What is the relationship? ForceMassAcceleration Force  Constant  Mass  Constant  Acceleration  Constant."— Presentation transcript:

1 Newton’s Second Law The Mathematical One

2 What is the relationship? ForceMassAcceleration Force  Constant  Mass  Constant  Acceleration  Constant  Think of a race car! Force is generated by the engine Acceleration is how fast it increases it’s speed off the line

3 Student Race… Let’s hit the hall!

4 What are the relationships? Force is proportional to acceleration ▫If F  then a  (Mass constant) Force is proportional to mass ▫If F  then m  (Acceleration constant) Acceleration is inversely proportional to mass ▫If a  then m  (Force constant)

5 Scenario 1: Talk to your partner about how these relationships apply to: You are late for school and your mom pulls up behind a transport truck. What would the impact be if it was: Empty? Full?

6 Scenario 2: Talk to your partner about how these relationships apply to: You have 2 cars that have the same mass – one car has the engine of a Ford Focus, the other has the engine of a Ford Mustang. Assume both overall masses are the same

7 Newton’s Second Law When a net force acts on an object, the object accelerates in the direction of the net force. F NET = ma Unit Analysis: 1N = 1 kg ۰ m/s 2

8 Example # 1: A student is bowling with her friends. She gives a 7.0 kg bowling ball an acceleration of 5.0 m/s 2 [forward]. Calculate the net force she exerted on the ball. ā = 5.0 m/s 2 m = 7.0 kg F net = mā = (7.0 kg)(5.0 m/s 2 [forward]) = 35 N [forward]

9 Example 2: A motorcycle and driver have a combined mass of 280 kg. They accelerate from 7.0 m/s [E] to 34 m/s [E] in 4.2 s. What is the net force on the motorcycle and driver?

10 m = 280 kg v 1 = 7.0 m/s [E] v 2 = 34 m/s [E] Δt = 4.2 s Solve for ā first ā = v 2 – v 1 Δt = 34 m/s [E] – 7.0 m/s [E] 4.2 s = 6.4 m/s 2 [E] F net =ma = (280 kg)(6.4 m/s 2 [E]) = 1.8 x 10 3 N [E]

11 Newton’s Second Law and Motion Day 2 – The Real Math

12 Equations Motion Forces BOTH!!!

13 Example 1: A rocket ship has engines that can produce a force of 1.8x10 5 N [up]. If the ship has a mass of 1.3x10 4 kg. a)What will the acceleration of the ship be? b)What will it’s speed be after 10s?

14 Example 1: A driver approaches an intersection at a velocity of 14 m/s [forward] when the light turns amber. The driver applies the brakes to get the maximum stopping force. The car has a mass of 1500 kg, and the force of friction between the tires and the road is 1.1 x 10 4 N. Ignoring the driver's reaction time calculate: a.) the maximum deceleration of the car b.) the minimum stopping time c.) the minimum stopping distance

15 Example 2: An elevator, including passengers, has a mass of 600 kg. When leaving the first floor, it accelerates upward at 2.0 m/s 2. What force is the cable exerting on the elevator?


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