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Physics Chapter 5

Physics Chapter 5. Work and Energy. Work. Work - ( if force is constant ) – is the product of the force exerted on an object and the distance the object moves in the direction of the force. W = F || d.

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Physics Chapter 5

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  1. Physics Chapter 5 Work and Energy

  2. Work • Work - (if force is constant) – is the product of the force exerted on an object and the distance the object moves in the direction of the force. • W = F||d

  3. Work is a scalar quantity and can be positive or negative, but you don’t have to indicate direction/angle in your answer. • Unit is Joule (N•m) • The object must MOVE in the DIRECTION (positive or negative) that the force is exerted on THE OBJECT. • Holding a flag in the air (no work on the flag) • Walking forward while holding the flag in the air (no work on flag)

  4. Work and Direction of Force • If force is applied at an angle, then the work must be computed using the component of force in the direction of the motion. • Ex: you are pushing a lawnmower…

  5. Work and Direction of Force F • Force (parallel to ground) doing work is Fcos • Work (W) = Fcosd= Fdcos • Work that the grass/lawn is doing on the mower = Ffrictiond  Fcos d Ff

  6. Example How much work is done on a vacuum cleaner pulled 4.0 meters by a force of 45 N at an angle of 30.0 degrees above horizontal? 45N 30 Fcos Here’s the equation: W = Fd = F cos d = Fdcos W = Fd = F cos d = 45Ncos30X 4.0 m = 156 Joules

  7. D  mg  mgx Wgrav = mg sin d   Or Wgrav = mgd if lifted vertically

  8. Gravity Example • How heavy is a barrel that is lifted 45 meters vertically using 670 joules of work? • Formula for work done against gravity. • Wagainst grav = -Wgrav =-mgd • 670J = -mgd • m=-670J/(gd) = -670/(-9.81m/s2X45m) = 1.5 kg

  9. Practice 5A page 170 and Section Review on page 171

  10. Happy Thursday! • Please take out 5A and section review for me to check.

  11. Kinetic Energy • Energy - an object has energy if it can produce a change in itself or its surroundings. • Kinetic Energy - energy due to the motion of an object. • KE = 1/2mv2

  12. Doing Work to Change Kinetic Energy • Kinetic Energy is related to the amount of work done on the object. • Work done equals the KE gained by the object.

  13. Example pg. 173 • A 7.0 kg bowling ball moves at 3.00m/s. How much kinetic energy does the bowling ball have? How fast must a 2.45 g ping pong ball move in order to have the same kinetic energy as the bowling ball?

  14. Known, unknowns and formulas • mbb=7.0kg • Vbb=3.00m/s • mppb=2.45g = 0.00245kg • KEbb = 1/2mbbvbb2 =KEppb = 1/2mppbvppb2

  15. Solve for Kinetic Energy of Bowling Ball • KEbb = 1/2mbbvbb2 =(½(7.0kg)(3.00m/s)2= =31.5 J = 32J

  16. Solve for vppb • KEbb = 32J=KEppb = 1/2mppbvppb2 • vppb= 32J/(1/2mppb) • vppb= 32J/[(1/2)(0.00245kg)] vppb=160m/s

  17. Example 2 • What is the speed of a 0.20 kg baseball that has a Kinetic Energy of 390Joules? • m= 0.20 kg, KE = 390J, v=? • KE = ½ mv2 v= √2KE/m • v= √2(390J/0.20kg) = 62m/s

  18. Example 3What is the mass of a satellite traveling at 92 km/h with a kinetic energy of 240,000 J • KE = 240,000, v=92km/h????? • 92km/hX1000/3600=26m/s • KE=1/2mv2 , m=2KE/v2 • m=2(240,000J)/(26m/s)2 = 710kg

  19. 5B page 174

  20. Effect of Doing Work • As positive work is done on an object (lifting a box), energy is transferred to that object (it can now fall and crush something). • As negative work is done on an object (lowering a box), energy is transferred from the object to the one lowering the box.

  21. The Work-Energy Theorem • The net work done on an object is equal to its change in kinetic energy. • Wnet = KEf - KEi =  KE • If the net work is positive, the kinetic energy increases. • If the net work is negative, the kinetic energy decreases.

  22. Example • On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.20 m/s. How far does the sled move if the coefficient of kinetic friction between the sled and the ice is 0.10?

  23. Known, unknowns and formulas • ms=10.0kg • Vs=2.20m/s • k = 0.100 • KEf = 0 • Fnet = Fk…the only force working once the sled is kicked, but it is NEGATIVE in direction!! • Fnet=-Fk=-k Fn = k mg • Wnet = -Fkd = KEf – KEi • -k mgd = -KEi = -1/2msvs2

  24. Solve for d • d =-1/2msvs2 /-(k mg)= -1/2(10.0kg)(2.20m/s)2/-(0.10)(10.0kg)(9.81m/s2) =2.46m It still travels a positive distance, but it slows down as it is traveling so its change in kinetic energy, friction and acceleration are negative.

  25. Practice 5C

  26. Potential Energy - energy stored in an object because of its state or position.

  27. g is positive for potential energy! • Gravitational potential energy depends on an object’s position above a zero level. • PEg = mgh PEg = mghf - mghi = Work done • Potential energy can be stored in bent or stretched objects=Elastic Potential Energy • PEelastic=1/2kx2 • x = distance compressed or stretched • k = spring constant

  28. What is the Potential Energy of a 24.0 kg mass 7.5 m above the ground • PEg = mgh = (24.0)(9.81)(7.5m) = 1.8X103 J

  29. What is the Potential Energy of a spring stretched 5.0 cm that has a spring constant of 79 N/m? • PEE = ½ kx2 = ½ (79N/m)(.05m)2 = = 0.099J

  30. Total Potential energy =Elastic Potential Energy + Gravitational Potential Energy • PETotal=1/2kx2 + mgh

  31. Example • A 70.0 kg stuntman is attached to a bungee cord with an unstretched length of 15.0 m. He jumps off a bridge spanning a river from a height of 50.0 m. When he finally stops, the cord has a stretched length of 44.0 m. Treat the stuntman as a point mass, and disregard the weight of the bungee cord. Assuming the spring constant of the bungee cord is 71.8 N/m, what is the total potential energy relative to the water when the man stops falling?

  32. Known, unknows and formulas • ms=70.0kg • hi=50.0m, hf=?... • hf= 50m-44m=6.0m • Li= 15.0m • Lf = 44.0m • L = 29.0m =x • k=71.8N/m • PETotal=1/2kx2 + mghf • PETotal=1/2(71.8N/m)(29m)2+ (70.0kg)(9.81m/s2)(6.0m) = 3.43X104J

  33. Swing problem… (hyp)cos = adj Height above zero point = hyp-adj hyp  adj swing

  34. Problems 5D, page 180 and section review page 181

  35. Potential and Kinetic Energy • When a ball is thrown into the air, the kinetic energy you give the ball is transferred to potential energy and then back into the same amount of kinetic energy -- Total Energy is constant. • ETOTAL = KET + PET

  36. Mechanical Energy • Mechanical Energy (ME) is the sum of KE and ALL forms of PE associated with an object or group of objects. • ME = KE + PE

  37. Law of Conservation of Energy • States that within a closed, isolated system, energy can change form, but the total amount of energy is constant. • (Energy can be neither created nor destroyed).

  38. MEi=MEf • KEi + PEg,i+ PEE,i =KEf + PEg,f+ PEE,f ½ mvi2+mghi + ½ kxi2= ½ mvf2+mghf+½kxf2

  39. Example 1 • Starting from rest, a child slides down a frictionless slide from an initial height of 3.00 m. What is his speed at the bottom of the slide? Assume he has a mass of 25 kg.

  40. hi = 3.00m, hf = 0.00m Vi=0 m=25.0kg g=9.81m/s2 Vf=?

  41. Use the Conservation of Mechanical Energy Formula (no PEE) • KEi + PEg,i =KEf + PEg,f • mghi = ½ mvf2 • vf=(ghi )/(½ ) • vf= (9.81)(3.00m)/(1/2) • =7.67m/s

  42. Example 2. A brick with a mass of 2.1 kg falls from a height of 2.0 m and lands on a relaxed spring with a constant of 3.5 N/m. How far will the spring be compressed? Assume final position is 0.0 m and final KE is 0. No KE initial or final Given: m = 2.1 kg hi = 2.0 m g = 9.81 m/s2 k = 3.5 N/m x Find: Formula: Solution:

  43. Practice 5E

  44. Conservation of Energy Lab

  45. Power • Work is not affected by the time it takes to perform it. • Poweris the rate of doing work. • P=Wnet/t = Fnet,||d/t OR P= Fnet,|| v • Remember… Wnet = KE • F = ma • Friction/Resistance is a force that needs to be included in Fnet calculations • Units are in watts (W) • One Watt = 1 Joule/second • Kilowatt = 1000 watts

  46. Power Example 1. How long does it take for a 1.70 kW steam engine to do 5.6 x 106 J of work? (assume 100% efficiency.) Given: P = 1.70 x 103 W W = 5.6 x 106 J Find: Dt Formula: Solution:

  47. Power Example 2. How much power must a crane’s motor deliver to lift a 350 kg crate to a height of 12.0 m in 4.5 s? Given: m = 350 kg h = 12.0 m Dt = 4.5 s g = 9.81 m/s2 P Find: P=Wnet/t = Fnetd/t Formula: Solution:

  48. Power Example 3. A 1.2 x 103 kg elevator carries a maximum load of 900.0 kg. A constant frictional force of 5.0 x 103 N retards the elevator’s upward motion. What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s? Hint: The key to this problem is to find the required net force required in the vertical

  49. Power Fnet = Fw elevator + Fw load + Ffriction = g(me + ml) + Ff Fnet = 9.81 m/s2(1.2 x 103 kg + 900.0 kg) + 5.0 x 103 N Fnet = 25 601 N (without rounding)

  50. Power Given: Fnet = 25 601 N n = 3.00 m/s Find: P Formula: Solution:

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