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Physics 203 College Physics I Fall 2012

Physics 203 College Physics I Fall 2012. S. A. Yost. Chapter 6 Part 2. Conservative and Non-Conservative Forces Potential Energy. Announcements. Today is the last class on Chapte r 6. The first homework set HW06A was due today, the second, HW06B, is due Tuesday.

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Physics 203 College Physics I Fall 2012

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  1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 6 Part 2 Conservative and Non-Conservative Forces Potential Energy

  2. Announcements • Today is the last class on Chapter 6. The first homework set HW06A was due today, the second, HW06B, is due Tuesday. • The Exam on Chapters 4 – 6 will be next Thursday. • Tuesday: Read Ch. 7, sec. 1 – 3 on momentum and impulse. This isn’t on the exam. • A problem set HW07A will be posted, due the Tuesday after the exam. Set HW07B will come later.

  3. Quiz: Question 1 • Which of the following is nota conservative force? • A) Gravitational Force • B) Electric Force • C) Friction • D)Elastic (spring) force

  4. Quiz: Question 2 • If the potential energy of a roller coaster at points 1 and 2 on a track are U1and U2, how are the kinetic energies K1and K2 related at these points? • A) K1 – K2 = U1 – U2 • B) K2 – K1 = U1 – U2 • C) K1 + K2 = U1 + U2

  5. Quiz: Question 3 • Suppose two dart guns are identical and have equally stiff springs , except that dart gun 2 compresses the spring twice as far as gun 1 when the dart is loaded. If the guns are fired vertically upward, the dart fired from gun 2 will • A) Rise about as high as the dart from gun 1. • B) Rise about twice as high as the dart from gun 1. • C) Rise about 4 times as high as the dart from gun 1.

  6. Quiz: Question 4 • If a cart having mass 10 kg is accelerated for 60 seconds using a 12 Watt motor, starting at rest, how fast is it going at the end of this time? • A) 4.0 m/s • B) 6.0 m/s • C) 12 m/s • D) 24 m/s • E) 36 m/s

  7. Work by a Spring • When a spring is stretched or compressed, it exerts a restoring force is given by Hooke’s Law, • F = – k x. • k = spring stiffness constant. x x → F x = 0

  8. 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

  9. 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

  10. 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 does the spring do on the ball? m L • L = 0.040 m • k = 12,000 N/m • W = ½ kL2= 9.60 J.

  11. Spring Gun • 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

  12. Gravitational Work • Suppose I ride a ski-lift to the top of a hill and then ski back down to the starting-point. What is the total work done on me by gravity? Wup = -mgh Wdown = mgh Wtotal = 0 h

  13. Gravitational Work • This is always true for gravity. The work done along any closed path is zero. • When this is true for all paths, a force is said to be conservative. h

  14. Work by a Spring • What about a spring… • If you pull it from x1 to x2, and then let it go back to x1, has the spring done any work on your hand? • NO • F(x) = – kx x x 1 x 2 W1→2 = ½ k (x12 – x22) W2→1 = ½ k (x22 – x12) = – W2→1

  15. Friction • If I push a box weighing 10 N across a 5 mfloor, with coefficient of friction m= 0.10, and then push it back to where I started, how much work did friction do? • You can answer this… • Ff = 1.0 N, Wf = - 5 N each way. Fp Round trip work = - 10 N. Ff

  16. Conservative Forces • Gravity and the spring force are conservative. Friction is not. • In one dimension, any force that depends only on position must be conservative. • Friction also depends on the direction of motion, so it is not conservative.

  17. Potential Energy • For any conservative force, the amount of work done on an object going from a reference point r0 to an arbitrary point rdepends only on rand r0, not the path. • We can define a function • U(r) = U(r0) – Wr→r • U(r) is called the potential energy. → → → → → → → → 0 →

  18. Gravitational Potential Energy • Gravitational potential energy depends only on the height. The reference height h = 0 can be chosen arbitrarily. • Then U(h) = mgh This is the work you would have to do against gravity to reach the height h. h h=0

  19. Elastic Potential Energy • For a spring, it is convenient to take the equilibrium position to be x = 0, and set U(0) = 0. • Then the potential energy is • U(x) = ½ kx2. x x=0 x This is the work you would have to do against the spring to pull it a distance x from its equilibrium position.

  20. Energy Conservation • When an object moves from point 1 to point 2, a conservative force does work • W1→2 = U1 – U2. • The work-energy theorem says the change in kinetic energy is K2 – K1 = W1→2. • Then K2 – K1 = U1 – U2so • K2 + U2 = K1 + U1. • Energy Conservation: The total energy K + U is constant.

  21. Energy Conservation • When only conservative forces act on a system, energy is conserved. • All of the fundamental forces in the universe are conservative. • That means that if you keep track of the total energy in a closed system, it can never increase or decrease – it can only change form.

  22. Two water slides have the same length, but are shaped different. Who is going faster at the bottom of the slides? A) Paul B) Kathleen C) No difference Water Slide

  23. Who gets to the bottom first? A) Paul B) Kathleen C) No difference Water Slide

  24. 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. • Horsepower: 1 hp = 746 W.

  25. Example • How long will it take 50 hp motor to pull a 120 kg sled 100 m up a hill, if the coefficient of kinetic friction is m = 0.10, and the elevation increases by 20 m on the way up? Assume a constant slope.

  26. Example • First find the work done by the motor. • Assume the net work is zero. • Wm + Wf = DU = mgh • = 2.35 ×104 J. h = 20 m

  27. Example m= 0.10 m = 120 kg x = 98 m • Work done by friction? • Wf = –Ffd • Ff = mN, N = mg cosq • Ff = m mgcosq = – mmg (cosq) d = – mmg x = – 1.15 ×104 J → N d = 100 m → Ff h = 20 m θ → mg x =√ 1002 – 202 m θ

  28. Example • Wm = mgh – Wf = 2.35 ×104 J – (– 1.15 ×104 J ) • = 3.50 ×104J. • Time: t = Wm/Pm • Pm = 5.0 hp (746 W/hp) = 3730 W. • t = 9.4 s. h = 20 m

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