1 / 54

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.

devona
Download Presentation

Physics Chapter 5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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:

More Related