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PHYSICS 231
Lecture 10: Too much work!
Remco Zegers
PHY 231
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60N
50 kg
(top view)
80N
Two persons are dragging a box over a floor. Assuming there
is no friction, what is the acceleration of the crate?
a) 1.2 m/s2
b) 1.6 m/s2
c) 2.0 m/s2
d) 2.8 m/s2
e) no acceleration whatsoever
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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
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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
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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 next lecture
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
h=35m
d=34m
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h
end
Consider the above rollercoaster (closed system). The cart
starts at a height h. What is its velocity v at the end?
Hint: consider the mechanical energy in the beginning and
the end.
a) v=2gh
b) v=gttrip c) v=(2gh) d) v=(2gh/m) e) v=0 m/s
kinetic energy: 0.5mv2
potential energy: mgh
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