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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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  2. Lecture 12 Energy and Energy Transfer 2 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan

  9. The friction Force The Normal Force 9 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan

  12. Work done by a varying force 12 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan

  20. Now use the Work-Energy Theorem: 20 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) 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 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11) General Physics 1, Lec 12, By/ T.A. Eleyan

  24. Additional Questions (Lecture 12) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  25. [1] The force acting on a particle varies as in Figure . Find the work done by the force on the particle as it moves (a) from x = 0 to x = 8.00 m, (b) from x = 8.00 m to x = 12.0 m, and (c) from x = 0 to x = 12.0 m. [2] A 4.00-kg particle is subject to a total force that varies with position as shown in Figure. The particle start from rest at x = 0. What is its speed at (a) x = 5.00 m, (b) x = 10.0 m, (c) x = 15.0 m? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  26. [3] A 0.300-kg ball has a speed of 15.0 m/s. (a) What is its kinetic energy? (b) What If? If its speed were doubled, what would be its kinetic energy? [4] A 40.0-kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find (a) the work done by the applied force, (b) the increase in internal energy in the box-floor system due to friction, (c) the work done by the normal force, (d) the work done by the gravitational force, (e) the change in kinetic energy of the box, and (f) the final speed of the box. [5] The electric motor of a model train accelerates the train from rest to 0.620 m/s in 21.0 ms. The total mass of the train is 875 g. Find the average power delivered to the train during the acceleration. [6] A 700-N Marine in basic training climbs a 10.0-m vertical rope at a constant speed in 8.00 s. What is his power output? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  27. [7] A 1500-kg car accelerates uniformly from rest to a speed of 10 m/s in 3s. Find (a) the work done on the car in this time, (b) the average power delivered the engine in the first 3s, and (c) the instantaneous power delivered by the engine at t = 2s. [8] A mechanic pushes a 2500kg car from rest to a speed v doing 5000J of work in the process.  During this time, the car moves 25m. Neglecting friction between the car and the road, (a) What is the final speed, v, of the car? (b) What is the horizontal force exerted on the car? [9] A 200kg cart is pulled along a level surface by an engine.  The coefficient of friction between the carte and surface is 0.4.  (a) How much power must the engine deliver to move the carte at constant speed of 5m/s?  (b) How much work is done by the engine in 3min? [10] A 65kg woman climbs a flight of 20 stairs, each 23 cm high.  How much work was done against the force of gravity in the process? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  28. [11] A skier of mass 70.0 kg is pulled up a slope by a motor driven cable. (a) How much work is required to pull him a distance of 60.0 m up a 30.0° slope (assumed frictionless) at a constant speed of 2.00 m/s? (b) A motor of what power is required to perform this task? [12] If an applied force varies with position according to F.= 3x3 - 5, where x is in m, how much work is done by this force on an object that moves from x = 4 m to x = 7 m? [13] A horizontal force of 150 N is used to push a 40-kg box on a rough horizontal surface through a distance of 6m.  If the box moves at constant speed, find (a) the work done by the 150-N force, (b) the work done by friction. [14] When a 4-kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.5cm. If the 4-kg mass is removed, (a) how far will the spring stretch if a 1.5-kg mass is hung on it, and (b) how much work must an external agent do to stretch the same spring 4.0 cm from its unstretched position? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  29. [15] A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of 1.75 m/s. (a) What is the average power of the elevator motor during this period? (b) How does this power compare with the motor power when the elevator moves at its cruising speed? [16] A 4.00-kg particle moves along the x axis. Its position varies with time according to x = t + 2.0t 3, where x is in meters and t is in seconds. Find (a) the kinetic energy at any time t, (b) the acceleration of the particle and the force acting on it at time t, (c) the power being delivered to the particle at time t, and (d) the work done on the particle in the interval t = 0 to t=2.00 s. [17] If a man lifts a 20-kg bucket from a well and does 6 kJ of work, how deep is the well?  Assume the speed of the bucket remains constant as it is lifted. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  30. Lecture 13 Potential Energy Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  31. There are two types of forces: • conservative (gravity, spring force) • All microscopic forces are conservative: • Gravity, • Electro-Magnetism, • Weak Nuclear Force, • Strong Nuclear Force • nonconservative (friction, tension) • Macroscopic forces are non-conservative, Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  32. Conservative Forces A force is conservative if the work it does on an object moving between two points is independent of the path taken.  work done depends only on ri and rf Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  33.  If an object moves in a closed path (ri = rf) then total work done by the force is zero. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  34. Nonconservative Forces  work done by the force depends on the path  non-conservative forces dissipate energy Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  35. Work Done by Conservative Forces Potential Energy:Energy associated with the position of an object. For example: When you lift a ball a distance y, gravity does negative work on the ball. This work can be recovered as kinetic energy if we let the ball fall. The energy that was “stored” in the ball is potential energy. Wc = -DU =-[Ufinal– Uinitial] Wc = work done by a conservative force DU = change in potential energy Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  36. Gravitational Potential Energy • Gravitational potential energy U = mg(y-y0),where, y = height • U=0 at y=y0 (e.g. surface of earth). • Work done by gravity: • Wg = -mg Dy = -mg (y- y0) Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  37. Spring Potential Energy Uf – Ui = - [Work done by spring on mass] Mass m starts at x=0 (Ui =0) and moves until spring is stretched to position x. WorkSpring = - ½ kx2 U(x) – 0 = - (-1/2 kx2) USpring(x) = ½ kx2 x = displacement from equilibrium position F=-kx x Area in triangle = - kx times increment in x = Work done by spring Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  38. Conservation of Energy • Energy is neither created nor destroyed • The energy of an isolated system of objects remains constant. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  39. Mechanical Energy (Conservative Forces) Mechanical energyE is the sum of the potential and kinetic energies of an object. E = U + K The total mechanical energy in any isolated system of objects remains constantif the objects interact only through conservative forces: E = constant Ef =Ei Uf + Kf = Ui+ Ki DU + DK = DE = 0 Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  40. Example: A ball of mass m is dropped from a height h above the ground, as shown in Figure (A) Neglecting air resistance, determine the speed of the ball when it is at a height y above the ground. (B) Neglecting air resistance, determine the speed of the ball when it is at the ground. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  41. Example:  A 0.5 kg block is used to compresses a spring with a spring constant of 80.0 N/m a distance of 2.0 cm.  After the spring is released, what is the final speed of the block? Solution Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  42. Example: A particle of mass m = 5.00 kg is released from point A and slides on the frictionless track shown in Figure. Determine (a) the particle’s speed at points B and C and (b) the net work done by the gravitational force in moving the particle from A to C. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  43. (a) the particle’s speed at points B Find the particle’s speed at points C then find the work from the relation Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  44. Work Done by Nonconservative Forces Nonconservative forces change the amount of mechanical energy in a system. Wnc = work done by nonconservative force Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  45. Problem: Children and sled with mass of 50 kg slide down a hill with a height of 0.46 m.  If the sled starts from rest and has a speed of 2.6 m/s at the bottom, how much thermal energy is lost due to friction (i.e. what is the work that friction does)?  If the hill has an angle of 20° above the horizontal what was the frictional force. Since vi = 0, and hf = 0, Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  46. The force done by friction is determined from; Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  47. Example: A child of mass m rides on an irregularly curved slide of height h = 2.00 m, as shown in Figure. The child starts from rest at the top. (A) Determine his speed at the bottom, assuming no friction is present. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  48. (B) If a force of kinetic friction acts on the child, how much mechanical energy does the system lose? Assume that vf =3.00 m/s and m = 20.0 kg. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  49. Example: A block having a mass of 0.80kg is given an initial velocity vA = 1.2m/s to the right and collides with a spring of negligible mass and force constant k = 50N/m, as shown in Figure. (A) Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision. Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

  50. (B) Suppose a constant force of kinetic friction acts between the block and the surface, with = 0.50. If the speed of the block at the moment it collides with the spring is vA = 1.2 m/s, what is the maximum compression xC in the spring? Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)

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