Physics 203 College Physics I Fall 2012

1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 6 Part 1 Work and Kinetic Energy

2. Announcements • Read Chapter 6 for next time. You can skip sec. 6-2. • Today we will discuss sec. 6.1, 6.3, and 6.10 on Work, Kinetic Energy, and Power. • There is a problem set on these sections due Thursday: HW3A • Next time, we will discuss sec. 6.4 – 6.9: Potential Energy, Energy Conservation, Non-Conservative Forces. • Problem set HW3B is due next Tuesday.

3. Newton’s Law and Orbits • First: an orbit problem… • Determine the mass of the sun using the properties of Earth’s orbit. (You can treat it as circular.) • Newton’s Gravitational law: F = G Ms Me / R2. • Newton’s 2nd Law: F = Mea = Mev2/R • Uniform circular motion: v = 2pR / T • G Ms Me / R2 = Me(2p/T)2 R • Ms = (2p/T)2 R3/ G

4. Newton’s Law and Orbits • Ms = 4p2R3/(GT2) • R = 1.50 × 1011 m • T = 1 year = 3.16 × 107 s • G = 6.67 × 10–11 Nm2/kg2 • The numbers give Ms = 2.00 × 1030 kg

5. Effect of Force over Distance • Applying a force to a particle over distance changes its speed in the direction of the force: vf2 – vi2= 2ax (1 dim) Multiply by ½ m:½ mvf2 – ½mvi2= max UseF = ma: ½ mvf2 – ½ mvi2 = Fx

6. The Work-Energy Theorem Definitions: Kinetic Energy =K = ½ mv2. Work = W =Fx. Units: Work = N m = Joules (J). Work-energy principle: The work done by the netforce on a mass causes a change in kinetic energy: DK = W

7. Work Requires Motion • Work is done only when there is motion. • W = Fxrequires both F and x to be nonzero for W to be nonzero. • You can push all day on a wall and get very tired, but if it doesn’t move, you did no work on it.

8. Question Suppose you apply a force Fp = 50 N to a box, which causes it to move at a constant speed through a distance of 10 m. What is the net work done on the box? A) 0.2 J B) 500 J C) 250 J D) 0 J E) 5 J Ff Fp = 50 N 10 m

9. Example Suppose you apply a force Fp = 60 N to a box of mass m = 15 kg initially at rest, with coefficient of kinetic friction mk= 0.3. Two forces act on the box: Fpand Ff = mkmg. Fp Fp x

10. Example • 1. How much work do you do on the box when it moves a distance x= 12 m? You do an amount of work Wp = Fp x = (60 N)(12 m) = 720 Nm = 720 J The pushing force is in the direction of motion, so the work is positive. Fp x

11. Example 2. How much work does friction do when you move the box 12 m? The force of friction is Ff = – mkmg = – (0.3)(15 kg)(9.8 m/s2) = –44 N The work done by friction is Wf = Ffx = (–44 N)(12 m) = –528 J Fp x

12. Example 3. What is the kinetic energy of the box after being pushed 12 m? The box was initially at rest, so the kinetic energy is the net work, K = Wp + Wf= 720 J – 528 J = 192 J 4. What is the speed of the box after being pushed 12 m? The kinetic energy isK = ½ m v2. v = √ 2 K/m = √ 2(192 J) / 15 kg = 5.1 m/s

13. Orbit • A satellite of weight mg is in a near-Earth circular orbit. How much work does gravity do on the satellite during each orbit? • A) mgRe • B) 2pmgRe • C) 0 • Think of the work-energy principle: DK = W. Re

14. Work in More than 1 Dimension Only the component of force in the direction of motion does work. A force perpendicular to the motion does no work! It can’t change the speed, so it doesn’t affect the kinetic energy. W =DK y → F W = (F cosq) x θ x

15. High Dive via Work/Energy • A diver jumps with initial speed v0 = 1.4 m/s. At what speed does she enter the water 5.0 m below? • Kf = K0 + Wwith K0 = ½ mv02, • Kf = ½ mvf2. • Gravity is a constant force in the -y direction, so W = - mg Dy = mgh. The x motion does not affect the work. v0 5.0 m vf

16. High Dive via Work/Energy • Kf = K0 + W • ½ mvf2 = ½ mv02 + mgh • vf = √v02 + 2gh • v0 = 1.4 m/s, g = 9.8 m/s2, h = 5.0 m. • vf= 10.0 m/s. v0 5.0 m vf

17. Bowling Balls • A ball-feeder lifts balls up a 1 m long ramp to a platform 0.5 m above the floor. How much force must the feeder arm exert to lift a ball weighing 50 N? • Enter the answer in Newtons. 1.0 m 0.5 m

18. Bowling Balls • The feeder does work Wp = FpL = F × 1.0 m. • Gravity does work Wg = - mgh= -(50N)(0.5 m) = - 25 N • Wpmust be at least25 J. • Fpmust be at least 25 N. 1.0 m 0.5 m

19. Power • Power is the rate of doing work: P = W/t. • If the force F acts in the direction of motion, then • P = Fv (instantaneous) • These are consistent because x = v t is the distance traveled, so • P = F v = F x/t = W/t.

20. Bowling Balls • If the ball feeder is powered by a 5 W motor, how many balls per minute can it lift, in continuous operation? • Each ball required Wp = 25 J of work. The motor supplies (5 W) (60 s) = 300 J in a minute. That is enough energy to lift 12balls. 1.0 m 0.5 m

21. Pulleys • How much force do I have to pull with to lift a block of mass M at a constant speed? • Solve it using work and energy. Fp M

22. Pulleys • Pulling a distance Lwith force Fpdoes work • Wp= FpL • The block moves up a distance ½ Lwhile lifted with a forceFLift = Mg. • Wp = FpL = WLift= ½ LMg • Fp = ½ Mg Fp L FLift ½ L M

23. Work by a Spring • When a spring is compressed or stretched, there is a restoring force given by Hooke’s Law, • F = – k x. • k = spring constant. F L x x = -L x = 0

24. Work by a Spring • Suppose a ball is placed in front of the spring. • If it is held at x = –L and then released, how much work does the spring do on the ball as it returns to its equilibrium position, launching the ball? m x x = -L x = 0

25. Work by a Spring • The work done by a changing force is the average force times the distance. • The force decreases from kLdown to 0 linearly, so the average force is F = ½kL. • W = F L = ½ kL2. F kL W = ½ k L2 0 x -L 0

26. Spring Gun • Suppose a ball of mass m = 250 gis pushed back an distance L = 4 cmon a spring of spring constant k = 120 N/cmand released. • How much work is needed to compress the spring? m L • L = 0.040 m • k = 12,000 N/m • W = ½ kL2= 9.60 J.

27. Spring Gun • When the spring is released, what is the launch speed of the ball? • Work-Energy Theorem: • K = ½ mv2 = W = 9.60 J. • m = 0.250 kg. • v = √2K/m = 8.76 m/s. m v