1 / 25

Physics 2011

Physics 2011. Chapter 6: Work and Kinetic Energy. Work. The Physics of Work. By strict definition, in order for work to be performed, a Net Force must be applied to a body, resulting in the Displacement of that body. Work = Force * Displacement = Newtons * Meters

Download Presentation

Physics 2011

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. Physics 2011 Chapter 6: Work and Kinetic Energy

  2. Work

  3. The Physics of Work • By strict definition, in order for work to be performed, a Net Force must be applied to a body, resulting in the Displacement of that body. Work = Force * Displacement = Newtons * Meters = Joules (Energy)

  4. Calculating Work from Vectors • Consider the Idiot pushing his girlfriend’s car in one direction while she steers in another: • The useful work is: • Thus the Scalar, Work, is a DOT PRODUCT:

  5. Work has a Sign • Work is calculated by finding the component of Force acting along the line of Displacement, but they may be in opposite directions. • ALSO, Work is W and Weight is w …..OK?

  6. Work is ENERGY • Work is the product of a Net Force and an accompanying displacement • A body under the influence of a Net Force is accelerating (F = ma) • An accelerating body is said to have increasing Kinetic Energy

  7. Kinetic Energy • A body with Mass, m, moving at velocity, v, has some ability to perform Work (For example, a bowling ball rolling down the alley can knock over pins) • This ability of a moving body to do work (Work is Energy) is quantified as: Kinetic Energy, K = ½ mv2 (Joules)

  8. Work-Energy • Positive Work on a Body INCREASES its Kinetic Energy • Negative Work on a Body DECREASES its Kinetic Energy • A body that gains K must increase in speed and a body that loses K must decrease in speed.

  9. gotta have POWER!!!! Power is the RATE of Work: i.e. Power is the change in work over some unit of time P = ΔW / Δt (Average Power) P = dW/dt (Instantaneous Power) Power is Joules/Seconds or Watts

  10. Review: Sum of Constant Forces Suppose FNET = F1 + F2 and the displacement is S. The work done by each force is: W1 = F1 r W2 = F2 r FTOT F1 r WNET= W1 + W2 = F1 r + F2 r = (F1+ F2) r WNET= FNET  r F2

  11. Review: Constant Force... W = Fr • No work done if = 90o. • No work done by T. • No work done by N. T v v N

  12. v1 v2 F x Work/Kinetic Energy Theorem: {NetWork done on object} = {change in kinetic energy of object} WF = K = 1/2mv22 - 1/2mv12 WF = Fx m

  13. j Work done by gravity: • Wg = Fr= mgrcos  = -mg y (remember y = yf - yi) Wg = -mg y Depends only ony ! m mg r  y m

  14. = Fr1+ Fr2+ . . . + Frn = F(r1+ r2+ . . .+ rn) =Fr = Fy r1 r2 r r3 Wg = -mg y rn Work done by gravity... Depends only ony, not on path taken! W NET = W1 + W2 +. . .+ Wn m mg y j

  15. v=0 v=0 v=0 H vf vi vp Free Fall Frictionless inclinePendulum Falling Objects • Three objects of mass m begin at height h with velocity 0. One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of a pendulum. What is the relationship between their velocities when they have fallen to height 0? (a)Vf > Vi > Vp(b) Vf > Vp > Vi (c) Vf = Vp = Vi

  16. v = 0 v = 0 v = 0 H vf vi vp Free Fall Frictionless inclinePendulum Solution Only gravity will do work: Wg = mgH = 1/2 mv22 - 1/2 mv12 = 1/2 mv22 does not depend on path !!

  17. F(x) x1 x2 dx Work done by Variable Force: (1D) • When the force was constant, we wrote W = F x • area under F vs. x plot: • For variable force, we find the areaby integrating: • dW = F(x) dx. F Wg x x

  18. dv Work/Kinetic Energy Theorem for a Variable Force dv F dx F dx dv dv dv v (chain rule) dx = = dx dt dx dt dv dx v dx v dv v22 v12 v22 v12

  19. 1-D Variable Force Example: Spring • For a spring, Hooke’s Law states: Fx = -kx. F(x) x1 x2 x relaxed position -kx F= - k x1 F= - k x2

  20. Spring... • The work done by the spring Wsduring a displacement from x1to x2 is the area under the F(x) vs x plot between x1and x2. F(x) x1 x2 x Ws relaxed position -kx

  21. Spring... • The work done by the spring Wsduring a displacement from x1to x2 is the area under the F(x) vs x plot between x1and x2. F(x) x1 x2 x Ws -kx

  22. Work & Energy • A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed position while momentarily coming to rest. • If the initial speed of the box were doubled and its mass were halved, how far x2 would the spring compress ? (a)(b) (c) x

  23. In the case of x1 Lecture 10, Act 2Solution • Again, use the fact that WNET = DK. In this case, WNET =WSPRING = -1/2 kx2 and K = -1/2 mv2 so kx2 = mv2 x1 v1 m1 m1

  24. Lecture 10, Act 2Solution So if v2 = 2v1 and m2 = m1/2 x2 v2 m2 m2

More Related