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					Kinetic Energy

  A P R I L 1 4 TH, 2 0 1 1
                What is Kinetic Energy

 Kinetic energy is the energy of motion. An object
    that has motion - whether it is vertical or horizontal
    motion - has kinetic energy.
   There are many forms of kinetic energy:
    vibrational - the energy due to vibrational motion),
   rotational (the energy due to rotational motion)
    translational (the energy due to motion from one
    location to another)

 The amount of translational kinetic energy (from
  here on, the phrase kinetic energy will refer to
  translational kinetic energy) that an object has
  depends upon two variables:
 the mass (m) of the object and the speed (v) of the

 The following equation is used to represent the
  kinetic energy (KE) of an object:
             KE = (½)mv2
             m = mass of object (kg)
             v = speed of object (m/s)
 This equation reveals that the kinetic energy of an
  object is directly proportional to the square of its
  velocity. That means that for a double in velocity, the
  kinetic energy will increase by a factor of four
                       KE cont.

 Kinetic energy is a scalar quantity; it does not have a
 Unlike velocity, acceleration, force, and momentum,
  the kinetic energy of an object is completely
  described by magnitude alone.
 Like work and potential energy, the standard metric
  unit of measurement for kinetic energy is the Joule.
                    Example 1.

 Determine the kinetic energy of a 1000-kg roller
 coaster car that is moving with a speed of 20.0 m/s.

                        EK = ½ mv2
                = ½ (1000kg)(20.0m/s)2
                   = 200 000 Joules
              = 2.00 x 105 Joules or 200 kJ
                     Example 2.

 If the roller coaster car in the above problem were
 moving with twice the speed, then what would be its
 new kinetic energy?

                 EK = ½ mv2
            = ½ (1000kg)(40.0m/s)2
            = 800 000 Joules
            = 8.00 x 105 Joules or 800 kJ
                     Example 3.

 Missy Diwater, the former platform diver for the
  Ringling Brother's Circus had a kinetic energy of
  15kJ just prior to hitting the bucket of water. If
  Missy's mass is 50 kg, then what is her speed?
EK = ½ mv2
15 000 J = ½ (50kg)v2
2(15000) = 50v2
  30000 = v2
      v =24.5m/s
              Work-energy theorem

 The net work done on an object is equal to its change
  in kinetic energy.
 Note that the work in the work energy theorem (from
  yesterday’s class) is the work done on an object by a
  net force – it is the algebraic sum of work done by all
       W = Ekf – Eki = ∆Ek
 *** So the change in kinetic energy is equal to the
  work done on or by an object***
                     Example 4.

 Calculate the velocity of a fist with a mass of 750g
 while being slammed into a board with a force of 50
 N over a distance of 35 cm.(watch out for units!)
EK = W = ½ mv2
  W = F d = ½ (0.75kg) v2
  50 N x 0.35 m = ½ (0.75kg) v2
  17. 5/ 0.375 = v2
      6.83 m/s = v
                     Example 5.

 A shotputter heaves a 7.26kg shot with a final speed
  of 7.50m/s. a. What is the kinetic energy of the shot?
  b. The shot was initially at rest, how much work was
  done on it to give it this kinetic energy?
a. Ek= ½ mv2 = ½ (7.26kg)(7.50m/s)2
                = 204 J
b. W = ∆Ek = Ekf – Eki
      = 204J – 0J = 204J

 Pg. 238 #’s 19,20,21
 Pg. 245 #’s 22,23,25

Jun Wang Jun Wang Dr
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