1 PHYSICS 231 Lecture 10: Too much work!

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Transcript 1 PHYSICS 231 Lecture 10: Too much work!

PHYSICS 231
Lecture 10: Too much work!
Remco Zegers
Walk-in hour: Thursday 11:30-13:30
Helproom
PHY 231
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WORK
Work: ‘Transfer of energy’
Quantitatively: The work W done by a constant force
on an object is the product of the force along the
direction of displacement and the magnitude of
displacement.
W=(Fcos)x
Units: =Nm=Joule
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An example
n
n
T
fk
Fg
T
=45o
x
Fg
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opposite
The work done by the person on the suitcase: W=(Tcos45o)x
The work done by Fg on the suitcase: W=(Fgcos270o)x=0
The work done by n on the suitcase: W=(Fgcos90o)x=0
The work done by friction on the suitcase: W=(fkcos180o)x=uknx
The work done by the suitcase on the person: W=(Tcos225o)x
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Non-constant force
W=(Fcos)x: what if Fcos not constant while covering x?
Example: what if  changes while
dragging the suitcase?
Area=A=(Fcos)x
Fcos
Fcos
W=(A)
The work done is the same as the area under the graph
of Fcos versus x
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Power: The rate of energy transfer
Work (the amount of energy transfer) is independent of time.
W=(Fcos)x … no time in here!
To measure how fast we transfer the energy we define:
Power(P)=W/t (J/s=Watt) (think about horsepower etc).
P =(Fcos)x/t=(Fcos)vaverage
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Example
A toy-rocket of 5.0 kg, after the initial
acceleration stage, travels 100 m in 2 seconds.
What is the work done by the engine?
What is the power of the engine?
h=100m
W=(Fcos)h=mrocketg h=4905 J
(Force by engine must balance gravity!)
P=W/t=4905/2=2453 W (=3.3 horsepower)
or
P=(Fcos)v=mgv=5.0 9.81 100/2=2453 W
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Potential Energy
Potential energy (PE): energy associated with the position
of an object within some system.
Gravitational potential energy: Consider the work done by
the gravity in case of the rocket:
Wgravity=Fg cos(180o)h=-mgh=-(mghf-mghi)=mghi-mghf
=PEi-PEf
The ‘system’ is the gravitational field of the earth.
PE=mgh
Since we are usually interested in the change in gravitational
potential energy, we can choose the ground level (h=0) in a
convenient way.
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Another rocket
A toy rocket (5kg) is launched from rest and reaches
a height of 100 m within 2 seconds. What is the
work done by the engine during acceleration?
h=100m
h(t)=h(0)+v0t+0.5at2 100=0.5a22 so a=50 m/s2
V(t)=V(0)+at V(2)=0+50*2=100 m/s
Force by engine=(50+9.81)m=59.81*5=299.05 N
(9.81 m/s2 due to balancing of gravitation)
W=Fh=299.05*100=29905 J
Change in potential energy:
PEf-PEi=mgh(2s)-mgh(0)=4905-0=4905 J
Where did all the work (29905-4905=25000 J)go?
Into the acceleration: energy of motion (kinetic energy)
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Kinetic energy
Consider object that changes speed only
t=2s
x=100m
V=0
a) W=Fx=(ma)x … used Newton’s second law
b) v=v0+at so t=(v-v0)/a
c) x=x0+v0t+0.5at2 so x-x0=x=v0t+0.5at2
Combine b) & c)
Rocket:
2
2
d) ax=(v -v0 )/2
W=½5(1002-02)
Combine a) & d)
W=½m(v2-v02)
=25000 J!!
That was missing!
Kinetic energy: KE=½mv2
When work is done on an object and the only change
is its speed: The work done is equal to the change in KE:
W=KEfinal-KEinitial
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Conservation of mechanical energy
Mechanical energy = potential energy + kinetic energy
In a closed system, mechanical energy is conserved*
V=100
ME=mgh+½mv2=constant
5 kg
What about the accelerating rocket?
h=100m
At launch: ME=5*9.81*0+½5*02=0
At 100 m height: ME=5*9.81*100+½5*1002=29905
We did not consider a closed system! (Fuel burning)
* There is an additional condition, see slides 12,13,14
V=0
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Example of closed system
A snowball is launched horizontally from the top of a building
at v=12.7 m/s. The building is 35m high. The mass is 0.2 kg.
Is mechanical energy conserved?
V0=12.7 m/s
At launch:
ME=mgh+½mv2
=0.2*9.81*35+½*0.2*12.72=84.7 J
h=35m
d=34m
At ground:
ah=(v2-v0,ver2)/2 so v=26.2 m/s
V=(Vhor2+Vver2)=(12.72+26.22)=29.1
ME=mgh+½mv2
=0.2*9.81*0+½*0.2*29.12=84.7 J
ME is conserved!!
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Conservative forces
A force is conservative if the work done by the force when
Moving an object from A to B does not depend on the path
taken from A to B.
Example: gravitational force
Using the stairs:
Wg=mghf-mghi=mg(hf-hi)
h=10m
Using the elevator:
Wg=mghf-mghi=mg(hf-hi)
The path taken (longer or shorter)
does not matter: only the
displacement does!
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Non conservative forces
A force is non-conservative if the work done by the force
when moving an object from A to B depends on the path
taken from A to B.
Example: Friction
You have to perform more work
Against friction if you take the
long path, compared to the short
path. The friction force changes
kinetic energy into heat.
Heat, chemical energy (e.g battery or fuel in an engine)
Are sources or sinks of internal energy.
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Conservation of mechanical energy only
holds if the system is closed AND all
forces are conservative
MEi-MEf=(PE+KE)i-(PE+KE)f=0 if all forces
are conservative
Example: throwing a snowball from a
building neglecting air resistance
MEi-MEf=(PE+KE)i-(PE+KE)f=Wnc if some
forces are nonconservative.
Wnc=work done by non-conservative forces.
Example: throwing a snowball from a
building taking into account air resistance
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