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MECHANICS

General Physics I Part 2 (lecture 12-19). MECHANICS. 2009/2010. Instructor Tamer A. Eleyan. Lecture 12 Energy and Energy Transfer. 2. General Physics 1, Lec 12, By/ T.A. Eleyan. Work Done by a Constant Force.

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MECHANICS

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  1. General Physics I Part 2 (lecture 12-19) MECHANICS 2009/2010 Instructor Tamer A. Eleyan

  2. Lecture 12 Energy and Energy Transfer 2 General Physics 1, Lec 12, By/ T.A. Eleyan

  3. Work Done by a Constant Force When an object undergoes displacement under force then work is said to be done by the force and the amount of work (W) is defined as the product of the component of force along the direction of displacement times the magnitude of the displacement. where is the angle between the direction of the force and the direction of the motion.  3 General Physics 1, Lec 12, By/ T.A. Eleyan

  4. The SI unit of work is (N×m) which is given the name (Joule). 1 Newton×meter = 1 Joule • Work is a scalar • Work has only magnitude, no direction. If I push on a wall and the wall does not move (no displacement), the work is (0J) because the displacement is (0m). Note that if F is in the same direction as d, then  = 0, and W= Fd 4 General Physics 1, Lec 12, By/ T.A. Eleyan

  5. Work is done in lifting the box (why?) No work is done on the bucket to move horizontally (why?) 5 General Physics 1, Lec 12, By/ T.A. Eleyan

  6. Negative Work and Total Work Work can be positive, negative or zero depending on the angle between the force and the displacement. If there is more than one force, each force can do work. The total work is calculated from the total (or net) force: Wtotal = Ftotald cosq 6 General Physics 1, Lec 12, By/ T.A. Eleyan

  7. Example:  Suppose I pull a package with a force of 98 N at an angle of 55° above the horizontal ground for a distance of 60m. What is the total work done by me on the package? Solution: Note that (F cosq) is the component of the force along the direction of motion. (Along the direction of the package's displacement.) 7 General Physics 1, Lec 12, By/ T.A. Eleyan

  8. Problem:A force F = (6ˆi - 2ˆj) N acts on a particle that undergoes a displacement r = (3ˆi +ˆj) m. Find the work done by the force on the particle Example:A 0.23 kg block slides down an incline of 25° at a constant velocity.  The block slides  1.5 m.  What is the work done by the normal force, by gravity, and by friction? What is the total work done on the block? 8 General Physics 1, Lec 12, By/ T.A. Eleyan

  9. The friction Force The Normal Force 9 General Physics 1, Lec 12, By/ T.A. Eleyan

  10. Now what is the work done by each individual force. Work done by the Normal Force: Work done by the Frictional Force: Work done by the Gravitational Force Now, to finally determine the "net-Work" 10 General Physics 1, Lec 12, By/ T.A. Eleyan

  11. Force - Displacement Graph The work done by a force can be found from the area between the force curve and the x-axis (remember, area below the x-axis is negative): Constant force 11 General Physics 1, Lec 12, By/ T.A. Eleyan

  12. Work done by a varying force 12 General Physics 1, Lec 12, By/ T.A. Eleyan

  13. Example:A force acting on a particle varies with x, as shown in Figure .Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m. SolutionThe work done by the force is equal to the area under the curve from x= 0 to x= 6.0 m. This area is equal to the area of the rectangular section from 0 to 4 plus the area of the triangular section from 4 to 6. The area of the rectangle is (5.0 N)(4.0 m) = 20 J, and the area of the triangle is ½(2m)(5N)=5J. Therefore, the total work =25J 13 General Physics 1, Lec 12, By/ T.A. Eleyan

  14. Problem: An object is acted on by the force shown in the Figure. What is the work from 0 to 1.00m 14 General Physics 1, Lec 12, By/ T.A. Eleyan

  15. Work Done by a Spring Where k is a positive constant called (the spring constant) 15 General Physics 1, Lec 12, By/ T.A. Eleyan

  16. Kinetic Energy An object in motion has kinetic energy: m = mass v = speed (magnitude of velocity) The unit of kinetic energy is Joules (J). Kinetic energy is a scalar (magnitude only) Kinetic energy is non-negative (zero or positive) 16 General Physics 1, Lec 12, By/ T.A. Eleyan

  17. Work-Energy Theorem The net (total) work done on an object by the total force acting on it is equal to the change in the kinetic energy of the object: Wtotal = DKE = KEfinal - KEinitial 17 General Physics 1, Lec 12, By/ T.A. Eleyan

  18. Example : How much work does it take to stop a 1000 kg car traveling at 28 m/s? Solution: Problem: A 0.600-kg particle has a speed of 2.00 m/s at point (A) and kinetic energy of 7.50 J at point (B). What is (a) its kinetic energy at (A)? (b) its speed at (B)? (c) the total work done on the particle as it moves from A to B? 18 General Physics 1, Lec 12, By/ T.A. Eleyan

  19. Example: A 58-kg skier is coasting down a slope inclined at 25° above the horizontal. A kinetic frictional force of 70 N opposes her motion.Near the top of the slope the skier's speed is 3.6 m/s. What is her speed at a point which is 57 m downhill? Solution: the net-work = Wf+Wg : 19 General Physics 1, Lec 12, By/ T.A. Eleyan

  20. Now use the Work-Energy Theorem: 20 General Physics 1, Lec 12, By/ T.A. Eleyan

  21. Example: A 65-kg bicyclist rides his 10.0-kg bicycle with a speed of 12 m/s. • How much work must be done by the brakes to bring the bike and rider to a stop? • How far does the bicycle travel if it takes 4.0 s to come to rest? • What is the magnitude of the braking force? Solution: • Friction = only horizontal force • Wnet = Kf -Ki <0 • Wnet = 0 – (1/2) m vi2 • Wnet = - (1/2) (10+65) (12)2 • Wnet = - 5400 kg m 2 /s 2 = -5400 J 21 General Physics 1, Lec 12, By/ T.A. Eleyan

  22. Find acceleration first • v = v0+at • 0 = v0 + at, • a= -v0/t • a = -12/4 = -3 m/s2 • x= x0+v0 t+(1/2)at2 • x= 0 + (12)(4) + (1/2)(-3)(4)2 • x= 48 -24 = 24 m • Braking force • Wnet = Fnet d = FFriction d • FFriction = Wnet /d = (-5400)/(24) = -225 N 22 General Physics 1, Lec 12, By/ T.A. Eleyan

  23. Power Power is defined as the rate at which work is done. The SI unit of power is "watts" (W). Power can also be written as; Whenever you want to determine power, you must first determine the force and the velocity or the work being done and the time. 23 General Physics 1, Lec 12, By/ T.A. Eleyan

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