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Newton’s 2 nd Law

Newton’s 2 nd Law. Vocabulary: Newton’s 2 nd Law weight. “ Gonconda ” (1953) Rene Magritte. This topic can be found in your textbook on pp . 130-132. Watch this.

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Newton’s 2 nd Law

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  1. Newton’s 2nd Law • Vocabulary: • Newton’s 2nd Law • weight “Gonconda” (1953) Rene Magritte This topic can be found in your textbook on pp. 130-132.

  2. Watch this The distance of this race is ¼ mile which is 400. m. Using the times of each car displayed at the end, calculate the average acceleration of each car. Click here to watch video

  3. Average acceleration The SMART car: a = 3.71 m/s2 The Honda S2000: a = 4.07 m/s2 For simplicity, let’s round both to 4 m/s2. mass = 680 kg mass = 1360 kg

  4. Average acceleration The SMART car: a = 4 m/s2 The Honda S2000: a = 4 m/s2 • Which car required more force to accelerate it at 4 m/s2? Why? • So is net force directly or inversely related to the mass it accelerates? mass = 680 kg mass = 1360 kg

  5. Result • Force and mass are directly related. More mass requires more force to give it the same acceleration. • Isaac Newton determined the relationship which we now call it Newton’s Second Law: or Acceleration is directly related to the net force acting on an object and inversely related to the mass of the object.

  6. The SMART car: a = 4 m/s2 The Honda S2000: a = 4 m/s2 Application • So what net force was required to accelerate each car? mass = 680 kg mass = 1360 kg

  7. The SMART car: a = 4 m/s2 The Honda S2000: a = 4 m/s2 Average acceleration mass = 1360 kg mass = 680 kg F = ma F = (680kg)(4m/s2) F = 2720 kg m/s2 F = 2720 N F = ma F = (1360kg)(4m/s2) F = 5440 kg m/s2 F = 5440 N Notice that double the mass required double the force to achieve the same acceleration!

  8. Important Points • A net force on an object will always result in the acceleration of the object. • This is right in line with Newton’s first law! • If there is no net force, velocity will be constant! • More mass requires more force to achieve the same acceleration. • In terms of Newton’s 2nd Law, why do objects with different masses have the same acceleration in free fall?

  9. Practice 1 A 0.50kg roadrunner’s legs exert a force of 2.3 N. If the air resistance is 1.7 N, what is the acceleration of the roadrunner? (Draw a FBD.) Fn Fnet = m a (2.3 N + -1.7N) = (0.50 kg) a a = 1.2 m/s2 2.3 N 1.7 N weight

  10. Practice 2 What net force needs to be applied on a 4162 lb (1888 kg) Bugatti Veyron to bring the car from its max speed of 268 mi/hr (120. m/s) to a stop in 1740 ft (531 m)? We know that Fnet = ma. Notice that to solve for the force, we need the mass (which is given) and the acceleration (which is not given). We will need to use information in the problem to first solve for acceleration.

  11. Practice 2 What net force needs to be applied on a 4162 lb (1888 kg) Bugatti Veyron to bring the car from its max speed of 268 mi/hr (120. m/s) to a stop in 1740 ft (531 m)? G U E S S vi = 120. m/s vf = 0 m/s ∆x = 531 m a = ? vf2=vi2+2a∆x a = -13.6 m/s2 (0m/s)2=(120.m/s)2+2a(531m)

  12. Practice 2 What net force needs to be applied on a 4162 lb (1888 kg) Bugatti Veyron to bring the car from its max speed of 268 mi/hr (120. m/s) to a stop in 1740 ft (531 m)? (a = -13.6 m/s2) G U E S S Fnet = ma m = 1888 kg a = -13.6 m/s2 Fnet = ? Fnet = -25,700 N Fnet = (1888 kg)(-13.6 m/s2)

  13. Gravity Weight is a measure of the force due to gravity. If you recall, the acceleration due to gravity, g, is 9.8 m/s2. So we can use Newton’s second law to calculate the weight of an object: F = ma for gravity: Fg = weight = mg weight = mg WEIGHT IS A FORCE!!!

  14. Practice 3 What is the acceleration of 62 kg Pearl E. Gates at a certain point as she skydives if the air resistance force is 423 N? air resistance weight = mg

  15. Practice 3 What is the acceleration of 62 kg Pearl E. Gates at a certain point as she skydives if the air resistance force is 423 N? 423 N Fnet = 423 N + (-607.6N) = -187 N -(62 kg)(9.8N) = -607.6 N

  16. Practice 3 What is the acceleration of 62 kg Pearl E. Gates at a certain point as she skydives if the air resistance force is 423 N? Fnet = -187 N Fnet = ma -187 N = (62 kg) a a = -3.0 m/s2

  17. Newton’s 2nd Law Summary Equation Summary • Fnet = ma • w = mg

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