Transcript Mechanical Energy

Mechanical Energy
Gravitational Potential &
Kinetic Energy
Gravitational Potential Energy:
Note:

Stored energy of a raised object
Eg  mgh
Gravitational
Potential
Energy (J)
Same
formula as
W = Fd
h  h final  hinitial
Mass of
object (kg)
Gravity
(N/kg)
How much potential energy does a 61.2kg
boy have if he is sitting on his bed, 0.5m
above the floor?
m = 61.2 kg
g = 9.8 N/kg
∆h = 0.5 m
Eg  mgh
Eg  (61.2)(9.8)(0.5)
Eg  299.88J
A 0.04kg rubber ball drops from a
height of 5m to the ground &
bounces back to a height of 3m.
a) How much potential energy does the ball
lose on the trip down?
m = 0.04 kg
g = 9.8 N/kg
∆h = 5 m
Eg  mgh
Eg  (0.04)(9.8)(5)
Eg  1.96J
b) How much potential energy does the ball
regain on the trip back up?
Eg  mgh Eg  (0.04)(9.8)(3) Eg  1.176J
Kinetic Energy:
Energy of an object due to its motion
 Result of work having been done to the
object
2
k

1
E  mv
2
Kinetic
Energy (J)
Mass of
object (kg)
Speed of
the object
(m/s)
What is the kinetic energy of a 6kg curling
stone sliding at 4m/s?
m = 6 kg
v = 4m/s
1
2
Ek  mv
2
1
2
Ek  (6)( 4)
2
Ek  48J
What is the speed of a 5.44kg shotput if its
kinetic energy is 68J?
m = 5.44 kg
Ek = 68J
1
2
Ek  mv
2
1
2
68  (5.44)( v)
2
v  5m / s
A 0.5kg rubber ball is thrown into
the air. At a height of 20m above
the ground, it is traveling at 15m/s.
What is the ball’s Ek & Eg?
m = 0.5 kg
v = 15m/s
So ball has
m = 0.5 kg
154.25J of
g = 9.8 N/kg
total energy
h = 20 m
1
Ek  mv 2
2
Eg  mgh
1
Ek  (0.5)(15) 2
2
Eg  (0.5)(9.8)(20)
Ek  56.25J
Eg  98J
BUT….total energy (ET)
still the same
throughout
Eg = high
Ek = zero
Eg = high
Ek = low
Eg = lower
Ek = higher
Eg = higher
Ek = lower
Eg = zero
Ek = high
Total Mechanical Energy (ET):

Energy can be transferred or transformed,
never lost  Law of Conservation of
Energy
ET  Eg  Ek *If friction
negligible
If
friction is not negligible then….
ET  Eg  Ek  EFriction