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Physics 1501: Lecture 13

Physics 1501: Lecture 13. Announcements HW 05: Friday next week Midterm 1: Monday Oct. 3 Covers Chapters 1-5 Practice test + solutions on web HW solutions: on the web. Office hours today Topics Review Work & Energy, Power Potential energy Conservative & non-conservative forces.

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Physics 1501: Lecture 13

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  1. Physics 1501: Lecture 13 • Announcements • HW 05: Friday next week • Midterm 1: Monday Oct. 3 • Covers Chapters 1-5 • Practice test + solutions on web • HW solutions: on the web. • Office hours today • Topics • Review Work & Energy, Power • Potential energy • Conservative & non-conservative forces

  2. Definition of Kinetic Energy : The kinetic energy of an object of mass (m) moving at speed (v) is: K = 1/2 m v2 Work Kinetic-Energy Theorem: Review Definition of Work: Work (W) of a constant force Facting through a displacement ris: W = F.r = Frcos  = Frr

  3. Work & Power: • Consider the following, • But, • So, Z3 GLC

  4. Work Done Against Gravity • Consider lifting a box onto the tail gate of a truck. m h The work required for this task is, W = F · d = (mg)(h) (1) W = mgh

  5. N F mgsinq mg mgcosq Work Done Against Gravity m h q To push the box with constant speed, F = mgsinq The length of the ramp is h/sinq So the work done is, W = Fd = (mgsinq)(h/sinq) W = mgh Same as before !

  6. Work Done by a Spring Fs Dx Force from the spring is Fs = -kx, Displacement is x. W =  F dx = - kx dx W = - 1/2 kx2 Remember these two results for Chapter 7.

  7. V = 30 m/s 5% Work and Power • What is the power required for a car (m=1000 kg) to climb a hill (5%) at v=30m/s assuming the coefficient of friction m = 0.03 ? Car

  8. v=30 m/s N f 5% m g  j i (1002+52)1/2 5 100 Solution P =F·v = f·v +m g·v • FBD f = -|N| i = - mg cos  i g = - g sin  i - g cos  j P= -  mg cos  v - mg sin  v = - mg v (  cos  + sin  ) = - mg v (0.03 100 + 5)/(1002+52)1/2 = - (1000 kg)(10 m/s2)(30 m/s)(0.08) = - 23 970 W ~ -24 kW The car needs to spend that power to get up the hill

  9. Chap.7: New Topic - Potential Energy • Consider a ball at some height above the ground. No Velocity Some Velocity

  10. New Topic - Potential Energy • Consider a ball at some height above the ground. What work is done in this process ? (Work done by the earth on the ball) W = F. Dx W = mgh cos(0) W = mgh h

  11. New Topic - Potential Energy • Consider a ball at some height above the ground. Before the ball falls it has the potential to do an amount of work mgh. We say the ball has a potential energy of U = mgh. By falling the ball loses its potential energy, work is done on the ball, and it gains some kinetic energy, W = K = 1/2 mv2 = -DU = mgh h

  12. New Topic - Potential Energy • What if the path it follows to the ground is different ? h/2 h 45 h/2 45

  13. mgsinq mg mgcosq New Topic - Potential Energy • What if the path it follows to the ground is different ? Force of gravity F1 = + mgsinq Distance is d1=(+h/2)/sinq W1 = F1 d1 = mgh/2 y q F2= -mgsinq d2 =(-h/2)/sinq h/2 h 45 W2 = (-mgsinq)(-h/2sinq) = mgh/2 h/2 45 W = W1 + W2 = mgh

  14. New Topic - Potential Energy • How much work is done when the ball has fallen halfway ? h h/2

  15. Lecture 13,ACT 1Work Done by Gravity • The air track is at an angle of 30 degrees with respect to horizontal. The cart (with mass 1 kg) is released. It bounces back and forth on the track. It falls 1 meter down the track, then bounces back up to its original position. How much total work is done by gravity on the cart when it reaches its original position. 1 meter 30 degrees A) 5 J B) 10 J C) 20 J D) 0 J

  16. Going down: • Wd = F.Dx • Wd = [-mg sin(30)] [-d] =mgd/2 d = 1 m mg 30 degrees • Going up: • Wu = F.Dx • Wu = [-mg sin(30)] [d] = - mgd/2 d = 1 m mg 30 degrees Lecture 13,ACT 1Solution • Do this problem in two steps, going down and going up • Total: • W = Wd + Wu • W = mgd ( 1/2 – 1/2 ) = 0 The answer is D) 0 J Note: W didn’t depend on path

  17. Some Definitions • Conservative Forces- those forces for which the work done does not depend on the path taken, but only the initial and final position. • Potential Energy- describes the amount of work that can potentially be done by one object on another under the influence of a conservative force • W = -DU only differences in potential energy matter.

  18. ò W=F.dr = - U U2 r2 r2 ò U = U2 - U1 = - W=-F.dr r1 U1 r1 Potential Energy • For any conservative force F we can define a potential energy function U in the following way: • The work done by a conservative force is equal and opposite to the change in the potential energy function. • This can be written as:

  19. A Conservative Force : Spring • For a spring we know that Fx = -kx. F(x) x1 x2 x relaxed position -kx F= - k x1 F= - k x2

  20. F(x) x1 x2 x Ws Ws -kx What is the Work done by the Spring... • The work done by the spring Wsduring a displacement from x1to x2 is the area under the F(x) vsx plot between x1and x2. kx1 kx2

  21. A Non-Conservative ForceFriction • Looking down on an air-hockey table with no air, Path 2 Path 1 For which path does friction do more work ?

  22. Path 2 Path 1 A Non-Conservative Force Since |W2|>|W1| the puck will be traveling slower at the end of path 2. Work done by a non-conservative force takes energy out of the system. W1 = -mmg d1 W2 = -mmg d2 since d2 > d1, -W2 > -W1

  23. Lecture 13,ACT 2Work/Energy for Conservative Forces • The air track is again at an angle of 30 degrees with respect to horizontal. The cart (with mass 1 kg) is released 1 meter from the bottom and hits the bumper with some speed, v1. You want the cart to go faster, so you release the cart higher. How high do you have to release the cart so it hits the bumper with speed v2 = 2v1? 1 meter 30 degrees A) 1 m B) 2 m C) 4 m D) 8 m

  24. Lecture 13,ACT 2Work / Energy • Case 2 Work done W = mgL2/2; (Note DU = -mgDy = -mgh2) Work energy theorem, W = DK 1/2 mgL2 = 1/2 m(v22-0) 1/2 mgL2 = 1/2 m((2v1)2-0) L2 = 4v12/g = 4h1 • Case 1 Work done W = mgL1/2; (Note DU = -mgDy = -mgh1) Work energy theorem, W = DK 1/2 mgL1 = 1/2 m(v12-0) L1 = v12/g 1 meter 30 degrees C) 4 m

  25. Lecture 13,ACT 3Work/Energy for Non-Conservative Forces • The air track is once again at an angle of 30 degrees with respect to horizontal. The cart (with mass 1 kg) is released 1 meter from the bottom and hits the bumper with some speed, v1. This time the vacuum/ air generator breaks half-way through and the air stops. The cart only bounces up half as high as where it started. How much work did friction do on the cart ? 1 meter 30 degrees A) 2.5 J B) 5 J C) 10 J D) –2.5 J E) –5 J F) –10 J

  26. Lecture 13,ACT 3Work / Energy for Non-Conservative Force • Start is at top, v=0. Finish is half way up when v=0. • Velocity starts and ends at zero. • Total Work done on the cart must be zero. Work done by gravity Wg = mgDy = mgL/4; Work done by friction must cancel this, WT = Wg + Wf = 0 Wf = - Wg Wf = -mgL/4 = -(1kg)(10m/s2)(1m)/4 = -2.5J 1 meter 30 degrees D) -2.5 J

  27. Conservation of Energy • If only conservative forces are present, the total energy (sum of potential and kinetic energies) of a systemis conserved. E = K + U E = K + U = W + U = W + (-W) = 0 using K = W using U = -W E = K + U is constant!!! • Both K and U can change, but E = K + U remainsconstant.

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