1 / 9

WORK

WORK. Work changes the energy of a system by applying an external force through a displacement ( x). F. . Fcos =F //. x. Δ x. W = F • r = Fxcos = F // x. Units: Nm = Joule (J) W = Work F // = The force parallel to the displacement x = displacement. F (N). Work (Energy).

mimi
Télécharger la présentation

WORK

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. WORK

  2. Work changes the energy of a system by applying an external force through a displacement (x).

  3. F  Fcos=F// x Δx W = F•r = Fxcos = F// x Units: Nm = Joule (J) W = Work F// = The force parallel to the displacement x = displacement

  4. F (N) Work (Energy) x (m) Graphically work is equal to the area under the graph of force (F) vs. position (x).

  5. 20N 5m Example 1: Calculate the work done by the 20N force. W = Fx cos = (20N)(5m)(cos 0) = 100 Nm

  6. 20N 30 5m Example 2: Calculate the work done by the 20N force. W = Fx cos = (20N)(5m)(cos 30) = 86.6 Nm

  7. 20N 5m Fg Example 3: Calculate the work done by the 20N force. W = Fx cos = (20N)(5m)(cos 90) = 0 Nm

  8. 20N 5m 30 Example 4: Calculate the work done by the 20N force. W= Fx cos = (20N)(5m)(cos 0) = 100 Nm

  9. 5m 35kg Example 5: A rope applies a tension of 400N to a 35kg crate. Assuming the crate starts from rest find the velocity of the crate after being raised 5 meters. (Neglect friction) 400N

More Related