Transcript Work and Energy

Conservation of Energy
“with non-conservative forces”
Some Key Terms
Internal force: any force exerted on an object in the system due
to another object in the system.
External force: any force exerted by a object that is not part of
the system on an object within the system.
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Open system: a system that can exchange both
matter and energy with its surroundings.
Closed system: a system that can exchange energy with
its surroundings, but not with matter.
Isolated system: a system that does not exchange either
matter or energy with its surroundings.
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So far we have only looked at conservation of energy for
conservative forces.
Now we will look at what happens the total energy of a system
when a non-conservative force acts on the system.
The work done by non-conservative forces is equal to the
difference of the final mechanical energy and the initial
mechanical energy of a system.
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In short, the Work/Energy theorem tells us that when you
do work on a system you increase its final energy.
W  E
If a force does work on a system so that the initial energy is
different then the final energy, then the force must be a non
conservative force. (A.K.A. dose not conserve energy)
Wnc  E
Wnc  E f  Ei
If we now rearrange and solve for the initial velocity we get,
Ei  E f  Wnc
If there is no work done by a no conservative force then,
Ei  E f  0
Ei  E f
We are right back to conservation of energy.
We can now see that the Law of Conservation of Energy and
the Work/Energy Theorem are in actuality the same thing.
Where the work is done by a non conservative force.
W  E
Wnc  E
Wnc  E f  Ei
Ei  E f  Wnc
Example: A 65.0 kg skydiver stepped of a hot air balloon that is
500 m above the ground. After freefalling a short distance, she
deploys a parachute, finally reaching the ground of the velocity of
8.00 m/s
a) Find the gravitational potential energy of the skydiver,
relative to the ground, before she jumps.
3.19 x 105 J
b) Find the kinetic energy of the skydiver just before she
lands on the ground.
2.08 kJ
c) How much work was done by air friction.
-3.17 x 105 J
Example: A roller coaster car of mass of 200 kg is moving at a
speed of 4.00 m/s at point A in the diagram. This point is 15 m
above the ground. The car then heads down the slope toward
point B, which is 6 m above the ground. If 3.40 x 103 J of work
are done by friction between points A and B, determine the speed
of the car at point B.
12.5 m/s
Do
Practice Problems Page 308 (pdf 41) #’s 18-23
(for #18 you will need to refer to the example problem on pg 305)
End of Chapter Review Pg 327 (pdf 44)
#’s 1, 2, 5, 22, 23, 26.
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