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 mv 2 (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

11. Review: Sum of Constant Forces FFF Suppose F NET = F 1 + F 2 and the S displacement is S. The work done by each force is: F  r W 1 = F 1   r F  W 2 = F 2   r F F TOT rrrr FF1FF1 FF2FF2 W NET = W 1 + W 2 F  r F  r = F 1   r + F 2   r FF  r = (F 1 + F 2 )   r F  r W NET = F NET   r

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

13. Work/Kinetic Energy Theorem: NetWork {Net Work done on object} = changekinetic energy {change in kinetic energy of object} W F =  K = 1 / 2 mv 2 2 - 1 / 2 mv 1 2 xxxx F v1v1 v2v2 m W F = F  x

14. Work done by gravity: F  r r W g = F   r = mg  r cos  = -mg  y (remember  y = y f - y i ) W g = -mg  y Depends only on  y ! j m rrrr gmggmg yy  m

15. Work done by gravity... Depends only on  y, not on path taken! m gmggmg yy j W NET = W 1 + W 2 +...+ W n rr  = F   r = F  y r1r1r1r1 r2r2r2r2 r3r3r3r3 rnrnrnrn F  rF  rF  r = F   r 1 + F   r 2 +... + F   r n F  r 1 rr n = F  (  r 1 +  r 2 +...+  r n ) W g = -mg  y

16. 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) (b) (c) (a) V f > V i > V p (b) V f > V p > V i (c) V f = V p = V i v=0 vivi H vpvp vfvf Free Fall Frictionless incline Pendulum

17. Solution Only gravity will do work: W g = mgH = 1 / 2 mv 2 2 - 1 / 2 mv 1 2 = 1 / 2 mv 2 2 does not depend on path !! v = 0 vivi H vpvp vfvf Free Fall Frictionless incline Pendulum

18. 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 area by integrating: –dW = F(x) dx. F x WgWg xx F(x) x1x1 x2x2 dx

19. Work/Kinetic Energy Theorem for a Variable Force F F dx dv dx dv v dv v22v22 v12v12 v22v22 v12v12 dv dx v dv dx v(chain rule) dt = =

20. 1-D Variable Force Example: Spring For a spring, Hooke’s Law states: F x = -kx. F(x) x2x2 x x1x1 -kx relaxed position F = - k x 1 F = - k x 2

21. Spring... The work done by the spring W s during a displacement from x 1 to x 2 is the area under the F(x) vs x plot between x 1 and x 2. WsWs F(x) x2x2 x x1x1 -kx relaxed position

22. Spring... F(x) x2x2 WsWs x x1x1 -kx The work done by the spring W s during a displacement from x 1 to x 2 is the area under the F(x) vs x plot between x 1 and x 2.

23. Work & Energy A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x 1 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 x 2 would the spring compress ? x (a) (b) (c) (a)  (b) (c)

24. Lecture 10, Act 2 Solution Again, use the fact that W NET =  K. x1x1 v1v1 so kx 2 = mv 2 m1m1 m1m1 In this case, W NET = W SPRING = - 1 / 2 kx 2 and  K = - 1 / 2 mv 2 In the case of x 1

25. Lecture 10, Act 2 Solution x2x2 v2v2 m2m2 m2m2 So if v 2 = 2v 1 and m 2 = m 1 /2