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					   Chapter 10 Glencoe Physics


Energy, Work, and Simple Machines
               2006
A. What is work?

• 1. Webster’s Definitions
   – a. “energy expended by a natural
     phenomenum”
   – b. “activity in which one exerts strength
     to do something”
2. Physics Definition of Work

• a. Product of a force acting on an object
  and the displacement of the object in the
  direction of the force
• b. W = F s
   – (1) work is a scalar quantity
   – (2) displacement is critical - w/o
     movement no work is done
c. SI unit of work is the Newton-meter or
      JOULE

   – (1) A joule is the work done by a force
     of one newton as it acts through a
     distance of one meter along the line of
     the force
   – (2) Also use the joule to measure energy
• d. Other units of work include the foot-
  pound, the erg, and the electron-volt (eV)
3. Examples of work calculations
• a. A person exerts a vertical force of 30.0N
  on a pail as the pail is carried a horizontal
  distance of 8.0 m at a constant speed. How
  much work does the 30N force do on the
  pail?
• You do no work, as the force is not along
  the lines of the distance moved. It would
  have to have been moved vertically for
  work to be done.
b. How much work do you do on an object of
weight mg as:
• (1) you lift it a distance of h meters straight
  up at constant speed; and, (2) you lower it
  through this same distance at a constant
  speed?
• To lift you pull up on the object with a force
  F = mg to counteract the weight
   – W=F s=mgh
• When you lower the object F and s are in
  opposite directions, so W = m g (-h) = -mgh
4. Work and the Direction of Force

• a. Only the component of the force in the
  direction of movement does work. The
  component of force perpendicular to the
  direction of motion does NO work.
• Key Point to Remember - What is critical
  is to determine the component of force in
  direction of motion
b. Note as seen in a previous example you can
have “negative” work

• (1) although work is a scalar quantity,
  negative work means the force is applied in
  the opposite direction from the object’s
  motion.
• (2) Work is also cumulative, so negative
  work will offset positive work.
• c. Example. Motion is in the x -direction

              Fy              F

                     
                             Fx
                    Motion

                         s


 w = Fx s = (F cos ) s = F s cos

  is the angle between the applied force’s
        direction and the direction of motion
d. Special Work cases


•   (1)  = 0o, cos = 1 and W = F s
•   (2)  =180o, cos = -1 and W = -F s
•   (3)  = 90o, cos = 0 and W = 0
•   (4) Note that if you push an object from
    point A to point B and then back to A, you
    would have no displacement, so net work
    done is zero.
 e. Example: A man cleaning his apartment
 pulls a vacuum cleaner with a force of 50N
 and angle of 30o to the horizontal. A frictional
 force of 40 N retards his motion, and the
 vacuum is pulled a distance of 3 meters.
• (1) Calculate the work done by the 50N
   force
• (2) Calculate the work done by the frictional
   force
• (3) Calculate the net work done on the
   vacuum by all forces acting on it.
• First draw a picture or free body diagram

                       FN
                                FA = 50N
         Ff=40N
                                 motion


                       W=mg
• W app force = (F cos)s = (50N cos30o)(3 m)=
              = 130 Joules
• Wf = (Ff cos) s = (40 cos 180o)(3m) =
              = -120 Joules
     (work is being done on the vacuum by
  rug)
• m g and FN do no work, so Wnet = Wf + WAF
               FN
                           FA = 50N
 Ff=40N
                    =30
                            motion

               W=mg
5. Kinetic Energy
• a. energy itself is the ability of an object to
   produce a change in itself or in the world
   around it
• b. KE is the energy an object has due to its
   motion
• c. KE = ½ mv2
    – Mass, m, is in kilograms
    – Velocity , v, is in m/s
    – KE is measured in Joules
6. The Work-Energy Theorem
• a. work is equal to the change in energy that
  an object undergoes
• b. work is the transfer of energy by
  mechanical means.
• c. first established by Joule in nineteenth
  century
• d. energy transfer can go both ways – from
  work on an object that increase the objects
  energy or work done by the object so that it
  transfers energy to its surroundings.
7. Force versus displacement Curve

• a. A graph which shows how force is applied to an
  object with respect to the object’s displacment
  from the starting point.
• b. Area under the curve is the work done by the
  force over that distance. Note that a negative
  force means it is being applied in the opposite
  direction from motion.
• c. Area is cumulative. Negative offsets positive
B. Power
• 1. Power is the work done in a unit of time
          Power = Work Done
                   time to do work

             P=W/t
• 2. Basic units - work is in joules, time is in
  seconds, and power is in watts
   – a. 1 kilowatt = 1000 watts
   – b. 1 horsepower = 746 watts
Please note that power is the timed rate at
which work is being done.

• 3. Examples
   – a. A motor lifts a 200 kg object straight
     up at a constant speed of 3.00 cm/sec.
     What power is being developed by the
     motor? Express your answer in watts and
     horsepower.
• Givens:
   – Constant velocity so Fmotor = m g
   – m = 200 kg
   – v = 3.00 cm/sec = 0.0300 m/s so in one
     second the object moves 0.0300 m
      • s = 0.0300 m
      • t = 1 sec
   – w = m g = 1960 N
• Find P. P = w / t and W = F s
• W = 58.8 J
• P = W / t = 58.8 J / 1 sec = 58.8 watts

  = 58.8 watts ( 1 HP/ 746 watts) =0.0788 HP
• NOTE
     Pavg = W / time = F (s)/ t = F vavg
      so power can be determined if you know
  a force causes an object to move a given
  distance at a constant speed
b. What average power is developed by an
800N physics teacher while running at a
constant speed up a flight of stairs rising 6
meters if he takes 8 seconds to complete the
climb?
• Givens:
   – F = w = 800 N
   – v = constant
   – s = 6 meters
   – t = 8 sec
• P=Fs/t
• P = 600 watts
c. A 50 kg student climbs a rope 5 m in length
at a uniform speed and stops at the top. (1)
What must her average speed have been in
order to match the power output of a 200 watt
bulb; (2) how much work did she do; and (3)
how long did it take her to climb the rope?
• Givens:
    – m = 50 kg
    –s=5m
    – constant velocity so F = m g
    – P = 200 watts
• P=Fv          so v = P / F
• v = P / F = 0.41 m/s

• W = F s = 490 N (5 m) = 2450 Joules

• t = W / P = 2450 Joules / 200 watts
      = 12.5 sec
C. Machines
 1. What is a machine?

• a. A device by which the magnitude,
  direction, or method of application of a
  force is changed so as to achieve some
  advantage.
• b. NOTE: A machine does not change the
  amount of work done and does not make the
  amount of work done less, just makes it
  easier for someone to do a job.
c. Simple machines

• lever - three classes determined by the
  relationship between the applied force, load,
  and fulcrum point
• pulley
• inclined plane (wedge)
• wheel and axle
• screw
   d. complex machines are made up of simple
       machines
2. Mechanical Advantage (MA)
• a. Use a machine NOT to lessen work but
  usually to lessen the force required to do the
  work, to make the output work easier to
  achieve
• b. Concerned with the work into a machine
  and the work out of the machine
                 Ideal Machine
      Wout                             Win
                  Win = Wout
c. In a real machine, however, this does not
occur and Wout < Win

• (1) Reason - input work energy is converted
  to heat due to friction
• (2)
      Wout                                 Win
                   Real Machine



                  Thermal Energy

                    Win > Wout
d. Mechanical Advantage = MA = Fout / Fin

• (1) Force ratio, so unit-less quantity
• (2) If MA > 1, then machine has increased
  the force you applied to the object moved
• (3) NOTE: the machine did not lessen the
  work output or input in any way.
e. Ideal Mechanical Advantage (IMA)

• (1) IMA = Din / D out
• (2) This is the Distance Ratio, and again it
  is unitless
• (3) Design a machine to have a given IMA,
  then work to maximize efficiency and MA
3. Efficiency

• (1) A ratio between MA and IMA expressed
  as a percent
• (2) Efficiency = MA / IMA x 100%

                = W out / W in x 100%
4. Examples:
• a. A bottle opener required a force of 35N to
  lift the cap of a bottle 0.90cm. The opener
  has an IMA of 8.0 and an efficiency of
  75%.
• (1) What type of simple machine is the
  basis for the opener?
• (2) What is the MA of the opener?
• (3) What force is applied to the bottle cap?
• (4) How far does the handle of the opener
  move?
• hardest part in any problem is to identify
  what items are what
• “what force is applied to the bottle cap?” is
  asking for the Force out of the machine
• And, “How far does the handle of the
  opener move?” is asking for the Distance in
• If you get confused, remember a machine is
  used to lessen the force in, and it does this
  by trading distance for force. So the
  distance in will always be greater than
  distance out and Force out > force in.
• Givens:
   – Fin = 35N
   – D out = 0.90 cm = 0.0090m
   – IMA = 8.0
   – Efficiency = 75%
• eff = MA / IMA x 100%
   – MA = eff x IMA = .75 (8.0) = 6.0
             note efficiency was converted to a
             decimal
   – MA = F out / F in
so F out = MA (F in) = (6.0)(35N) = 210N

• IMA = D in / D out

     so    D in = IMA (D out)
                = 8.0 (0.90cm)
                = 7.2 cm
b. A worker uses a pulley to raise a 225 N
carton 10 meters. A force of 110 N is
required and the rope is pulled 30 m.
•     (1) What is the MA of the system?
      (2) What is the IMA of the system?
      (3) What is the efficiency of the system?
• Givens:
   – F out = 225N      F in = 110N
   – D out = 10 m      D in = 30 m
• MA = 2.05 IMA = 3.0              eff = 68%
QUIZ - on separate sheet of paper

• What is the efficiency of a pulley if a force
  of 650 N acting through 15 meters is
  required to lift a 11,000 N crate up a
  distance of 0.750 meters?