1 / 27

Physics 203 College Physics I Fall 2012

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

vlad
Download Presentation

Physics 203 College Physics I Fall 2012

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 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

More Related