1 / 18

Chapter 5 Practice problems

Chapter 5 Practice problems. Net Force. Sample #1. Using Newton’s Second Law to find the net force on an accelerating object: …. What net force is required to accelerate a 1500 kg race car act at +3.00m/s 2 ?. Sample #1. Given: m = 1500 kg a =+ 3.00 m/s 2 Unknown: Force, F. Sample #1.

giacomo-franklin
Télécharger la présentation

Chapter 5 Practice problems

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 5 Practice problems Net Force

  2. Sample #1 • Using Newton’s Second Law to find the net force on an accelerating object: • …. What net force is required to accelerate a 1500 kg race car act at +3.00m/s2?

  3. Sample #1 • Given: m = 1500 kg a =+ 3.00 m/s2 • Unknown: Force, F

  4. Sample #1 • Basic equation: • F = ma

  5. Sample #1 • Answer: 4500 N force

  6. Sample #2 • Finding force when acceleration must be calculated:

  7. Sample #2 • An artillery shell has a mass of 55 kg. The shell is fired from a gun leaving the barrel with a velocity of + 770 m/s. The gun barrel is 1.5 m long. Assume that the force, and thus the acceleration is constant while in gun barrel. What is the force on the shell while it is in the gun barrel?

  8. Sample #2 • Given: M = 55 kg vf = + 770 m/s vi = 0 m/s • D = +1.5 m • Unknown: force, F • Basic equations: F = ma • Vf2 =vi2 + 2ad

  9. Sample #2 • Answer: 1.1 X 107 N

  10. Sample # 2 • Solution: = • A = (+770m/s2) - 0 • 2(1.5 m) • a = +2.0 X 105 m/s2 ( in the direction of the velocity) • F = 55 kg +2.0 X 105 m/s2

  11. Sample #3 • Calculating weight: • Find the weight of a 2.26 kg bag of sugar

  12. Sample #3 • g= 9.8 m/s2 • Equation: W = mg

  13. Sample #3 • Answer : 22.1 N

  14. Sample #4 • Sliding friction: • A smooth wooden block is placed on a smooth wooden tabletop. You find that you must exert a force of 14.0 N to keep the 40.0 N block moving at a constant velocity

  15. Sample #4 • A. What is the coefficient of friction for the block and table? • Given: FA = 14 N W = 40 N a = 0 • Unknown : Coefficient of friction = μ

  16. Sample #4 • Basic equation = F= ma Ff =FμFN

  17. Sample #4

  18. Sample #4 • Because a = 0 the force you exert, FA must be equal to Ff

More Related