Transcript Energy Loss

Energy IN = Energy OUT
Means ALL Energy
3.2.4 Work to Overcome Friction
Energy “Loss”
Energy can never by truly destroyed, but it can be
converted into non-mechanical forms (mainly
heat) that are not ‘useful’.
ET  PE  KE  Q
We now consider problems in which Q ≠ 0
Example #1 – Lifting a Load
In thejoules
“real world”
it takes
morein
energy
toalift
• 5200
of work
are done
lifting
50an
object object
than thea gain
in gravitational
PE would
kilogram
distance
of 10 meters.
suggest.
– Determine the amount of potential energy gained
by lifting this object.
ΔPE = mgΔh
ΔPE = (50 kg)(9.81 m/s2)(10 m)
ΔPE = 4905 J
Example #1 – Lifting a Load
• 5200 joules of work are done in lifting a 50
kilogram object a distance of 10 meters.
– How much energy must have been used to
overcome friction?
W = Fd = ΔET = 5200 J
ET = PE + KE + Q
5200 J = 4905 J + 0 + Q
Q = 295 J
Example #2 – Missing KE
world”object
potential
energy
notat
completely
• InA the
4.0“real
kilogram
begins
at isrest
the top
converted
when
objects
of an
inclineinto
thatkinetic
is 2.0energy
meters
above
the lose
height.
ground and slides down
to the bottom. The
object reaches a speed of 5.0 meters per
second at the bottom of the incline.
– Determine the total energy of the object at the top
ET = PE + KE + Q
of the incline.
ET = mgh + ½ mv2 + Q
ET = (4.0 kg)(9.81 m/s2)(2.0 m) + 0 + 0
ET = 78.48 J
Example #2 – Missing KE
– Determine the amount of energy used to overcome
friction as the block slides down the incline.
ET = PE + KE + Q
78.48 J = mgh + ½ mv2 + Q
78.48 J = 0 + ½ (4 kg)(5.0 m/s)2 + Q
Q = 28.48 J
Example #3 – Constant Velocity
world”object
we must
constantly
addaenergy
• InA the
2.0“real
kilogram
is pulled
along
flat to
a system
work) to
keep things
at a30
surface
at a (do
constant
speed
with amoving
force of
constant speed.
newtons.
– Calculate the amount of work done in moving the
object 4.0 meters.
W = Fd = ΔET
W = (30 N)(4.0 m)
W = 120 J
Example #3 – Constant Velocity
• A 2.0 kilogram object is pulled along a flat
surface at a constant speed with a force of 30
newtons.
– How much work must have been done against
friction in moving this object?
W = Fd = ΔET = 120 J
ET = PE + KE + Q
120 J = 0 + 0 + Q
Q = 120 J
End of 3.2.4 - PRACTICE