1 / 10

Forces

Forces. Hooke’s Law. Hooke:. In 1678, Robert Hooke published the results of his work on elasticity and the deformation of an elastic object. An object is said to be elastic if it returns to its original dimensions after the applied force is removed. Hooke’s Law:. low k = easy to stretch.

donnan
Download Presentation

Forces

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. Forces Hooke’s Law

  2. Hooke: • In 1678, Robert Hooke published the results of his work on elasticity and the deformation of an elastic object. • An object is said to be elastic if it returns to its original dimensions after the applied force is removed.

  3. Hooke’s Law: low k = easy to stretch • The amount of deformation of an elastic object is directly proportional to the force applied to deform it. F = kx F = force applied on spring (N) k = force constant of the spring (N/m) x = amount of deformation of the spring (m)

  4. Ex. A spring has a force constant of 0.02N/m. What force is exerted to stretch the spring 12cm? F = kx F = ? F = (0.02)(0.12) k = 0.02 N/m F = 0.0024N x = 0.12m

  5. Ex. A 31 N force stretches a bungee cord 43 cm. What is the force constant of the cord? F = kx F = 31N 31 = (?)(0.43) k = ? k = 72.09 N/m x = 0.43m

  6. Ex. A spring that has an original length of 0.125m is stretched to a total length of 0.173m when a force of 3.6N is applied to it. What is the force constant of this spring. F = kx 3.6 = (?)(0.048) F = 3.6N k = 75 N/m k = ? x= 0.048m 0.173 – 0.125 = 0.048m

  7. Ex. A 325 g mass is applied to a spring with a force constant of 5.0 N/m. How much does it stretch? F = kx F = mg F = ? F = 3.185 N m = 0.325kg k = 5.0 N/m g = 9.8 N/kg x = ? F = (0.325)(9.8) 3.185 = (5)(x) F = 3.185 N x = 0.637m

  8. Ex. in N A mass of 500g stretches a spring 8cm when it is attached to it. What additional weight would you have to add so that the spring would stretch 10cm? F = kx F = kx F = mg F = (0.5)(9.8) 4.9 = (k)(0.08) F = (61.25)(0.1) F = 4.9 N k = 61.25 N/m F = 6.125 N 6.125 N – 4.9 N = Fadditional F = 1.225 N

  9. Ex. A B The following graph describes the spring constant of three different springs. Force (N) C length (m) Which of the springs would be the easiest to stretch? • small slope = low k F = kx • low k = easy to stretch spring C is easiest to stretch k = slope

  10. Try: pg 183 #1 - 3

More Related