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Chapter 8: Work, Power & Energy. Conceptual Physics Mr. Latham. Section 8.1: Work. Impulse = Force x time Work = Force x distance W= Fd Vector quantity- quantity and direction Work is always in the same direction as the force Measures in Joules (derived unit) (N)(m) = J

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chapter 8 work power energy

Chapter 8:Work, Power & Energy

Conceptual Physics

Mr. Latham

section 8 1 work
Section 8.1: Work
  • Impulse = Force x time
  • Work = Force x distance
  • W=Fd
    • Vector quantity- quantity and direction
    • Work is always in the same direction as the force
    • Measures in Joules (derived unit)
    • (N)(m) = J
      • Larger values can be measured in kJ (kilo=1,000)
  • LabQuest “Force v Time live”
example problem pushing
Example Problem- Pushing
  • What is the work involved in moving a 10kg shopping cart a distance of 10m to the right with a force of 12N?
    • If no frictional forces are mentioned, then we need to multiply F and d.
  • W = Fd
  • W = (12N)(10m) = 120N m = 120 J
example problem lifting
Example Problem- Lifting
  • How much work is required to lift a 12kg box a distance of 5m with a force of 170N?
  • Now we need to concern ourselves with gravity. The box has an Fg of:
    • Fg = (12)(9.8) = 117.6N
  • This is pointing down. We are lifting with 170N of force pointing up. What is the net force?
    • SF = 170N+(-117.6N) = 52.4N
  • We now have a force we can use and a distance was provided.
    • W = (52.4N)(5m) = 262J
  • Generally speaking, when we lift something, work is equal to (mass)(gravity)(height).
    • W = m g h
example problem pushing with friction
Example Problem- Pushing with Friction
  • What if I try to move the shopping cart (10kg) 10m to the right with 12N of force, but there is 5N of frictional force trying to stop me?
  • Just as above, we need to find the net force and use that in our work equation.
    • SF = Fp + (-Ffr) = 12N - 5N = 7N
    • W = (7N)(10m) = 70J
  • We get less work out of our system because not all of it goes into moving. Some of it goes into making heat and sound of moving a shopping cart across the floor.
non constant force calculus
Non-Constant Force (Calculus)
  • If force is constant, then the work can be easily calculated. W=Fd
  • If the force varies, then we need to solve the problem graphically. The area under the curve of a graph will tell us work.
  • So we can divide it into smaller and smaller rectangles and find the area of each. (That’s calculus!)
lab 21 making the grade
Lab 21: Making the Grade
  • Purpose: To Investigate the force and distance involved in moving an object up an incline.
  • For calculating lifting work, path doesn’t matter, only change in height.
section 8 2 power
Section 8.2: Power
  • Power is the rate by which work is done. Rates always include a time element.
  • P = average power = work/time
    • = (energy transformed)/(time) = J/s = Watt (W)
  • P=W/t
  • Although we are used to this being an electrical term, it is derived from the conversion of mechanical energy into electrical energy.
  • The power of a horse refers to how much work it can do per unit of time. One horsepower (hp) is defined as 550ft lbs/s (British system), which equals 746Watts in SI.
example problem stair climbing
Example Problem: Stair Climbing
  • A 60 kg jogger runs up a long flight of stairs in 4.0s. The vertical height of the stairs is 4.5m. Estimate the jogger’s power output in watts and horsepower.
  • The work done is against gravity and equals W=mgh. To get power, we find work and divide by time.
  • P = W/t = mgh/t
    • = (60kg)(9.8m/s2)(4.5m)/(4.0s) = 660Watts
  • Since there are 746Watts in 1hp, we divide to find that the joggers’: 0.885 hp
other uses of power
Other uses of Power
  • Recall that P=W/t, and W = Fd
  • Substituting for W, we see that
    • P = Fd/t
    • P = F (d/t)
    • P = F v
      • Where v is the average speed of the object!
lab 22 muscle up
Lab 22: Muscle Up!
  • Purpose: To determine the power that can be produced by various muscles of the human body.
  • Incandescent bulb 100 Watts
  • TV 120 Watts
  • Washing Machine 325 Watts
  • Vacuum 750 Watts
  • Toaster 800 Watts
section 8 4 potential energy
Section 8.4: Potential Energy
  • Potential energy is the ability to do work.
  • Chemical potential energy
    • HCl in a glass jar
    • Battery, Oil
  • Nuclear potential energy
    • Radioactive Material
  • Mechanical potential energy
    • Springs
  • Gravitational potential energy
    • An object lifted above a surface
gravitational potential energy
Gravitational Potential Energy
  • We determined earlier that W = m g h
    • Where h = yf-yi
  • If we measure our height from some starting point on the surface, the h will be negative, giving us the gravitational potential energy used by the object.
  • PEg = m g h
  • The higher the object, the more PE it possesses. We always measure h from the surface it could hit (or the lowest possible surface, or some other relative point), not necessarily from sea level of Earth.
  • PhET “Energy Skate Park”
example problem roller coaster a
Example Problem- Roller Coastera
  • A 1000kg roller coaster car moves from point 1 to point 2 and then on to point 3. What is the PEg at point 2 and point 3 relative to point 1?
  • PEg1-2 = (1000kg)(9.8)(10m) = 9.8x104J
  • PEg1-3 = (1000kg)(9.8)(-15m) = -1.5x105J
    • (negative because it is below point 1)
example problem roller coaster b
Example Problem- Roller Coasterb
  • A 1000kg roller coaster car moves from point 1 to point 2 and then on to point 3. What is the change in PEg when the car goes from point 2 to point 3?
  • The DPEg would be PEgf – PEgi
    • = (-1.5x105J) - (9.8x104J) = -2.5x105J
  • So the PEg decreased by 2.5x105J
section 8 5 kinetic energy
Section 8-5: Kinetic Energy
  • Kinetic energy is the energy of motion.
  • Radiant Energy- light, x-rays, gamma rays, radio
  • Thermal Energy- heat
  • Sound- longitudinal waves (earthquake waves as well)
  • Electrical Energy- flowing electrons, lightening, electricity
  • Motion- movement of objects
    • The faster an object is moving or the larger it is (or both), the more kinetic energy it has.
  • The kinetic energy of an object at constant velocity
  • KE = ½mv2
example problem ke of baseball
Example Problem- KE of Baseball
  • A 145 g baseball is thrown so that it acquires a speed of 25m/s. What is its kinetic energy of the ball?
  • The kinetic energy is found with the KE formula
  • KE = ½mv2 = (½)(.145kg)(25)2 = 45J
  • Rev. Questions 7-9
section 8 6 conservation of energy
Section 8-6: Conservation of Energy
  • Non-conservative forces- are forces where it does matter what path you take. These would be things like air resistance, friction and tension.
  • Conservative forces- for work done where the path traveled doesn’t matter, like lifting something (PEg)
    • Therefore, PE can only be defined to conservative forces
  • When we are looking at conservative forces, all energy is conserved, so what comes from one type, goes completely into another.
  • DKE = DPE
  • KEf-KEi = PEf-PEi
conservation
Conservation
  • We now define a quantity ME as the total mechanical energy of our system.
    • The sum of the energies at any moment is
    • ME = KE + PE
  • If KEi + PEi = KEf + PEf
    • Then MEi = MEf = constant
  • Total mechanical energy of a system neither increases or decreases in any process. This is the principle of the conservation of mechanical energy.
  • To pull several of our equations together, KE, PE, and E
  • ME = KE + PE= ½mv2 + mgh
  • We can use this equation to find velocities, masses, height, etc.
example problem conservation
Example Problem- Conservation
  • Solve for one variable at a time and fill in the blanks.
    • PEg=mgh
    • KE=½mv2
    • ME=PEg+KE
law of conservation of energy
Law of Conservation of Energy
  • Work is done when energy is transferred from one object into another.
    • Pushing a shopping cart
    • Throwing a bowling ball
    • Firing an arrow from a bow
  • Law of Conservation of Energy: The total energy is neither increased nor decreased in any process. Energy can be transformed from one form to another, and transferred from one object to another, but the total amount remains constant.
  • HW Conceptual Development 8-1 - 8-3
section 8 7 machines
Section 8-7: Machines
  • Machine- device used to multiply the force or change the direction of the force
    • Based on the Conservation of Energy
  • Work input = Work output
    • Finput d = Foutput d
  • Mechanical Advantage
    • Force output/Force input = (Foutput)/(Finput)
    • Distance input/distance output = (dinput)/(doutput)
  • Lever- simplest machine
example problem see saw
Example Problem- See Saw
  • The portion you push down on a see saw is 1 meter above the ground and the other side is raised 0.25m. You push with 10N of force. What is the maximum weight you could lift?
  • Finput d = Foutput d
    • (10N)(1m) = (Foutput)(0.25m)
  • Foutput = (10N)(1m)/(0.25m)
  • Foutput = 40N
three classes of levers
Three Classes of Levers
  • Fulcrum between the force and the load
    • See-saw
  • Load between the force and the fulcrum
    • Lifting one side of a car with a long steel bar
  • Force between the load and the fulcrum
    • Forearm
pulleys
Pulleys
  • Pulley- a type of lever that can multiply forces
  • Fulcrum between the force and the load
    • First pulley
  • Load between the force and the fulcrum
    • Second pulley
  • Forces can be multiplied when using multiple pulleys
    • From the third image, when the rope is pulled 4m with a force of 100N, the 400N load is lifted 1m
    • Mechanical advantage = 400N/100N = 4
section 8 8 efficiency
Section 8-8: Efficiency
  • Efficiency- ratio of useful work output/work input
  • Cars are incredibly inefficient