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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

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Physics 2011

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  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

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