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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
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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)
Cont.
 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
object
Cont.
 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
direction.
 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
50
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
forces.
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
Homework
 Pg. 238 #’s 19,20,21
 Pg. 245 #’s 22,23,25