1. PHYS 152 Fall 2005 Thursday, Sep 22 Chapter 6 - Work, Kinetic Energy & Power

2. Exam #1 Thursday, September 29, 7:00 – 8:00 PM. Material: Chapters 2-6.3 Use black lead #2 pencil and calculator. Formula sheet provided. You may bring 1 extra sheet of handwritten notes (both sides) but no other materials Sample exams on Web (link on homepage) Room assignments - to be announced.

3. Pg 3 Supplemental Instruction Exam Review Session Tuesday Sept. 27 th 7-9pm RHPH 172

4. Pg 4 Reading Quiz: Sections 6.2 & 6.3 During a short time interval a particle moves along a straight line a distance During that time a constant force acted on the particle: The work done on the particle was a.0 b.4 c.8 d.16 e..

5. Pg 5 Two marbles, one twice as heavy as the the other, are dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has (a) as much kinetic energy as the lighter one (b) twice as much kinetic energy as the lighter one (c) half as much kinetic energy as the lighter one (d) four times as much kinetic energy as the lighter one (e) impossible to determine

6. Pg 6 Two marbles, one twice as heavy as the the other, are dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has (a) as much kinetic energy as the lighter one (b) twice as much kinetic energy as the lighter one (c) half as much kinetic energy as the lighter one (d) four times as much kinetic energy as the lighter one (e) impossible to determine

7. Pg 7 F r W = F  r If  = 90 o no work is done. Review: Constant Force... v N T v N No work done by N T No work done by T

8. Pg 8 Work-Kinetic Energy Theorem Work done by the net external (constant) force equals the change in kinetic energy NetWorkchangekinetic energy {Net Work done on object} = {change in kinetic energy of object}

9. Pg 9 First calculate the work done by gravity: g r r W g = mg  r = -mg  r Now find the work done by the hand: F r r W HAND = F HAND  r = F HAND  r Work done by Lifting Example: Lifting a book from the floor to a shelf gmggmg rrrr F F HAND v v = const a a = 0 floor shelf

10. Pg 10 Work done by Lifting Work/Kinetic Energy Theorem: W =  K When lifting a book from the floor to a shelf, the object is stationary before and after the lift: W NET = 0 W NET = W HAND + W g rr = F HAND  r - mg  r r = (F HAND - mg)  r W HAND = - W g gmggmg rrrr F F HAND v v = const a a = 0 floor shelf

11. Pg 11 Lifting vs. Lowering W HAND = - W g gmggmg rrrr F F HAND v v = const a a = 0 floor shelf gmggmg rrrr F F HAND v v = const a a = 0 floor shelf LiftingLowering r W g = -mg  r r W HAND = F HAND  r W HAND = - W g r W g = mg  r r W HAND = -F HAND  r

12. Pg 12 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 ) Work done by gravity... m gmggmg h j W NET = W 1 + W 2 +...+ W n W g = mg h rr = F  r = F  y

13. Pg 13 Work done by Variable Force: (1D) * When the force was constant, we wrote W = F  x area under F vs. x plot: F x WgWg xx F(x) x1x1 x2x2 dx * For variable force, we find the area by integrating: dW = F(x) dx.

14. Pg 14 dx Work/Kinetic Energy Theorem for a Variable Force F F dx dv v dv v22v22 v12v12 v22v22 v12v12 dv dx dv dx v(chain rule) dx dv dx v dt = =

15. Pg 15 Power is the rate at which work is done by a force P AVG = W/  t Average Power P = dW/dt Instantaneous Power The unit of power is a Joule/second (J/s) which we define as a Watt (W) 1 W = 1 J/s Power

16. Pg 16 The force on a particle of mass m is given by Choose the correct statement: a.The work done will be the same going from x=1 to x=2 as it is for going from x=0 to x=1. b.The work done will be the same going from x=1 to x=2 as it is for going from x=-1 to x=-2. c.The average power is the same as the instantaneous power. d.None of the above are correct.

17. Pg 17 Work done by a Spring *For a person to hold a spring stretched out or compressed by x from its unstretched length, it requires a force where k =spring constant measures the stiffness of the spring. Spring unstretched x=0 FsFs FpFp Person pulling FsFs FpFp Person pushing

18. Pg 18 Work done by a Spring The spring exerts a force (restoring force) in the opposite direction: where k =spring constant measures the stiffness of the spring. Spring unstretched x=0 FsFs FpFp Person pulling FsFs FpFp Person pushing Hooke’s law

19. Pg 19 1-D Variable Force Example: Spring *For a spring F x = -kx. ( Hooke’s Law) k =spring constant F(x) x2x2 x x1x1 -kx relaxed position F = - k x 1 F = - k x 2

20. Pg 20 1-D Variable Force Example: 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. F(x) x2x2 x x1x1 -kx relaxed position F = - k x 1 F = - k x 2 WsWs

21. Pg 21 1-D Variable Force Example: 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. F(x) x2x2 x x1x1 -kx WsWs

22. Pg 22 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. x v1v1 m1m1 m1m1

23. Pg 23 Work - Energy x1x1 v1v1 m1m1 m1m1 Use the fact that W NET =  K. so kx 2 = mv 2 In this case W NET = W SPRING = - 1 / 2 kx 2 and  K = - 1 / 2 mv 2 In the case of x 1

24. Pg 24 Work - Energy x2x2 v2v2 m2m2 m2m2 If the initial speed of the box were doubled and its mass were halved, how far x 2 would the spring compress ? (a) (b) (c)

25. Pg 25 Work - Energy x2x2 v2v2 m2m2 m2m2 If the initial speed of the box were doubled and its mass were halved, how far x 2 would the spring compress ? So if v 2 = 2v 1 and m 2 = m 1 /2

26. Pg 26 Example A person pulls on a spring. It requires a force of 75N to stretch it by 3 cm. How much work does the person do? If the person compresses the spring by 3 cm how much work does the person do? Calculate the spring constant: The work is The work to compress the spring is the same since W is proportional to x 2.

27. Pg 27 Example: Compressed Spring A horizontal spring has k=360N/m. (a) How much work is required to compress a spring from x=0 to x=-11 cm? (b) If a 1.85 kg block is placed against the spring and the spring is released what will be the speed of the block when it separates from the spring at x=0? mg x=0 F s =kx N x=-11

28. Pg 28 Example: Compressed Spring The work done to stretch or compress the spring is: In returning to its uncompressed length the spring will do work W=2.18J on the block. According to the work-energy principle the block acquires kinetic energy: