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Mechanical Energy

Mechanical Energy. Chapter 5. 5. 1 Work. WORK: When a force acts upon an object to cause a displacement (scalar quantity) Three quantities that must be known: Force acting on the object Displacement of the object Angle between the force and the displacement. 5.1 Work Continued.

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Mechanical Energy

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  1. Mechanical Energy Chapter 5

  2. 5. 1 Work • WORK: When a force acts upon an object to cause a displacement (scalar quantity) • Three quantities that must be known: • Force acting on the object • Displacement of the object • Angle between the force and the displacement

  3. 5.1 Work Continued • W=Fdcos • W is work (J or Nm or kgm2/s2) • F is force (N) (must have “a”) • d is displacement (m) • If goes in a circle, d=0. • cos is the angle between the force causing the displacement • Only the component parallel to displacement does work • Be very careful about identifying this angle!!

  4. 5.1 Identifying the Angle • A man carries a bucket down the hallway. • Force acting on the object is perpendicular to the object. • cos (90°)= 0 • Therefore, W=0. • The force of gravity is also perpendicular. • Gravity does no work on the bucket.

  5. 5.1 Signs for Work • Work can be positive or negative • If the boy lifts the box up, • F and x in same direction • =0° • Cos (0)=+1 • Work is positive • If the boy puts the box down, • F and x in opposite directions • =180° • Cos (180)=-1 • Work is negative

  6. 5.1 Practice • The box moves to the right for the following examples. Order these in the amount of work done, from most positive to most negative.

  7. 5.1 Practice • Dragging a dog 10.0 meters using 10.0 N force applied at 30°. • The force is 30° from the direction of the moving dog (Cos (30°)=0.866). • W=10.0 x 10.0 x 0.866=86.6 J • The force of friction (10.0 N) slows down a sliding book to a stop after 10.0 m. • The force is opposite the direction of the motion (Cos (180°)=-1). • W=10.0 x 10.0 x -1= -100. J • Using 10.0 N to pull a box 10.0 meters up a 30° incline. • The force is parallel to the direction the box is moving (Cos (0°)=1). • W=10.0 x 10.0 x 1=100. J

  8. 5.1 Practice (pg. 121) • An Eskimo pulls a sled loaded with salmon. The total mass of the sled and the salmon is 50.0 kg. The Eskimo exerts a force of 120. N by pulling the sled. • How much work does he do on the sled if the rope is horizontal to the ground, and he pulls the sled 5.00 m? • How much work does he do on the sled if he pulls at a 30.0 degree angle for the same distance? Remember weight is perpendicular to motion so it does NO WORK!

  9. 5.1 Wnet • Work requires a system of more than just one force. Frictional work is always involved (sometimes ignored). Total work done on an object is the sum of work done by each force. Wnet=F1dcos + F2dcos

  10. 5.1 Practice (pg. 122) • An Eskimo pulls a sled loaded with salmon. The total mass of the sled and the salmon is 50.0 kg. The Eskimo exerts a force of 120. N by pulling the sled. The coefficient of friction between the sled and snow is 0.200. • What is the force of friction if the rope is horizontal to the ground, and he pulls the sled 5.00 m? What is the work done by the friction? Total work? • What is the force of friction if he pulls at a 30.0 degree angled for the same distance? What is the work done by the friction? Total work? Remember weight is perpendicular to motion so it does NO WORK!

  11. Homework • Work Worksheet

  12. 5.2 Kinetic Energy • Kinetic Energy (J): the energy of motion • Depends on two variables: • Mass (kg) • Speed (m/s) • KE=1/2 mv2 • KE is directly proportional to the square of speed • KE is a scalar quantity

  13. 5.2 Practice A 7.00 kg bowling ball moves at 3.00 m/s. How much kinetic energy does it have? How fast must a 2.45 kg ping pong ball move in order to have the same kinetic energy?

  14. 5.2 Work-Energy Theorem Wnet=Fnetx Wnet=(ma) x V2=vi2+2ax Wnet=m() Wnet=1/2mv2+1/2mvi2 Wnet=KEf-KEi=KE

  15. 5.2 Work-Energy Theorem Continued Wnet=KEf-KEi=KE • This is only true if the change in KE is due entirely to the change in the object’s velocity. • +W=speeding up (KEf>KEi) • -W=slowing down (KEf<KEi)

  16. 5.2 Practice (pg. 124) • The driver of a 1000. kg car traveling at 35.0 m/s slams on his brakes to avoid hitting another car. After the brakes are applied, a constant friction force of 8000. N acts on the car. • What minimum distance should the brakes be applied to avoid a collision? • If the distance between the vehicles is only 30.0 m, at what speed would the collision occur?

  17. 5.3 Potential Energy • PE is the work that something “can” do or stored energy of an object • Only depends on the beginning and end points, not the path taken • Gravitational Potential Energy (GPE) • Spring Potential Energy (PEspring) • Chemical Energy • Nuclear Energy

  18. 5.3 Gravitational Potential Energy • Gravitational Potential Energy: the energy stored as the result of its HEIGHT • The energy is stored as a result of the gravitational attraction of the Earth for the object • GPE=mgh • g=9.8 m/s2 • GPE is measure in J

  19. 5.3 Gravitational Potential Energy Continued • GPE is dependent on TWO variables: • Mass (kg) • More mass, greater GPE • Height (m) • Higher the object, greater GPE • Usually the ground is considered a height of ZERO

  20. 5.3 Practice (pg. 128) A 60.0 kg skier is at the top of a slope. Calculate her potential energy at point A. Point B? Halfway?

  21. 5.3 Conservation of ME • Mechanical energy is the sum of potential and kinetic energies • For a falling object (or pendulum), the sum of KE and PE remains constant at all times (ME is conserved). ME=KE + PE KEi+ GPEi=KEf+GPEf 1/2mvi2+mghi=1/2mvf2+mghf

  22. 5.3 Practice What is the speed at Point C? What is the speed at Point D?

  23. 5.3 Practice (pg. 131) A diver of mass m drops from a 10.0 m board above the water’s surface. What is his speed at 5.00 m above the water’s surface? What is his speed when he hits the water?

  24. 5.3 Practice (pg. 131) A grasshopper launches itself at an angle of 45 degrees and rises to 1.00 m during the leap. With what speed did it leave the ground?

  25. 5.2 and 5.3 Homework Physics Chapter 5 Work and Energy Problems Sheet

  26. 5. 4 Spring Potential Energy • PEspring is the energy stored in elastic materials as the result of their stretching or compressing. • Ex: rubber bands, bungee chords, trampolines, springs, bow and arrows, etc. • The more they stretch, the more PEspring

  27. 5. 4 Hooke’s Law • A force is required to compress a spring. • Higher the force, the more compression • Fspring=-k.x • k is the spring constant (dependent on the material) • X is the distance of compression or stretch

  28. 5. 4 Hooke’s Law • Fspring=-kx • Doubling the displacement requires double the force • Equilibrium position: the natural position when no force is applied • F and x are in opposite directions • k is large for a stiff spring and small for a flexible spring

  29. 5.4 Formula for PEspring • PEspring=1/2 k.x2 • PE is measure in J • k is the spring constant • x is the amount of compression • If the spring is not stretched or compressed, then there is no PEspring (equilibrium position)

  30. 5.4 Conservation of ME (KE + GPE + PEspring)i = (KE + GPE + PEspring)f Sum of KE and PE are conserved Determine what energies you have initially Determine what energies you have at the end Solve for missing variable

  31. 5.4 Practice (pg. 136-137) • A block of mass 5.00 kg is attached to a spring with a spring constant of 400. N/m. The flat surface upon which the block rests is frictionless. If the block is pulled out to xi=5.00 cm and released, • What is the speed at the equilibrium point? • What is the speed at 2.50 cm?

  32. 5.4 Practice • A spring is compressed 12.0 cm from equilibrium and the PE stored is 72.0 J. • What is the spring constant in this case? • 10,000 N/m • 5,000 N/m • 1,200 N/m • None of the above

  33. 5.4 Practice A 70.0 kg stuntman is attached to a bungee cord of unstretched length 15.0 m. He jumps off a bridge over water from a height of 50.0 m. When he stops, the cord has stretched to 44.0 m. If the spring constant is 71.8 N/m, what is the total potential energy when the man stops? What is the GPE when the man stops? What is the PEspringwhen the man stops?

  34. 5.4 Hooke’s Law Lab Hypothesize if your spring will have a high or low spring constant based on the stiffness of the spring. Create a data table with mass (kg), force (N), and displacement (m) for at least six points. Create a graph with displacement on the x-axis and force on the y-axis. Draw a line of best-fit and calculate the slope. Determine the R2 value (standard deviation). I WILL HELP YOU DO THIS IN EXCEL ONCE YOU FINISH!!

  35. 5.4 Homework • Hooke’s Law Lab • Purpose was to find the spring constant for one of the springs. • Write your hypothesis • Describe steps used to find the spring constant. • Collect data (data table and graph). • Analyze data (what does the trendline and slope tell you). • Write a conclusion (R2 analysis).

  36. 5.2 Changes in Mechanical Energy • Conservative Forces: these DO NOT change total ME • Electrical, magnetic, Fg, or Fspring • Non-conservative Forces: these ARE capable of changing total energy because energy is dispersed in the form of heat or sound • Fapplied, Ffriction, Fair, Ftension, Fnormal

  37. 5.5 Conservation of Energy Conservation: overall energy remains constant MEi + Wext = Mef b/c Wext=0 for conservative forces MEi=MEf KEi + PEi = KEf + PEf Energy is transformed from one form to another but conserved overall

  38. 5.5 Examples of Conservative Forces Pendulums: Let go and it should rise to the same height every time (neglecting friction) Newton’s Cradle

  39. 5.5 Practice • Find how high you can jump • Find your mass (kg) and weight (N). • Use 1 lb=0.454 kg=4.448 N • Measure the maximum height of your jump. • Calculate your GPE at the maximum height. • Calculate your initial velocity from the ground. • How many Calories did you use for 1 jump? • Use 1 Calorie=4184 J

  40. 5.5 Non-Conservative Energy MEi + Wext = Mef KEi + PEi + Wext = KEf + PEf Work=ME • POSITVE change in ME • Energy is gained by the object • NEGATIVE change in ME • Energy is lost by the object

  41. 5.6 Power • Work can be done quickly or slowly • Power: the rate at which a certain amount of work is done (scalar quantity) • Power=work/time • Power is measure in Watts (W or J/s or Nm/s) • I Horsepower=746 Watts • For the same amount of work, power and time are inversely proportional

  42. 5.6 Power Continued ==Force Power=Force x Velocity A powerful machine is both strong AND fast

  43. 5.6 Practice • Identify which has more power. • A 120 J lift in 4 seconds. • A 200 J lift in 8 seconds. • During a physics lab, Jack and Jill ran up a hill. Jack is twice as massive as Jill, but Jill ascends the same distance in half the time. • Who did the most work? • Who delivered the most power?

  44. 5.6 Practice A 1500 kg car accelerates from rest to 10. m/s in 3.0 s. What is the work done and what is the power by the engine?

  45. Find Your Power Lab (Part 1) • All machines are described by a power rating, including people • FIND YOUR POWER!! • Quickly climb a flight of stairs. • Your weight=force • Height of staircase upward=d • Time=the time it takes for you to climb the stairs • Calculate the work required to climb the stairs. • How many Calories did you use? • Calculate your power rating for the stairs.

  46. Find Your Power Lab (Part 2) • Do 10 push-ups • Your weight=force • Length of your shoulder from the floor =d • Time it takes for you to complete 10 push-ups divided by 10 gives you the average time for 1 push-up • Calculate the work required to complete 1 push-up • How many Calories did you use? • Calculate your power rating for a push-up

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