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PHYSICS 231
Lecture 11: How much energy goes into
problems?
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Thompson R. (003)
Conti M. (001)
Zaran E. (002)
Whymer K. (002)
Aggarwal S. (?)
Chirio A. (001)
Webb A. (002)
Leboeuf J. (002)
Remco Zegers McKenzie A. (002)
Walk-in hour: Thursday 11:30-13:30 am
Helproom
PHY 231
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Previous lecture
• Work: W=Fcos(θ)∆x
• Power: P=W/∆t
• Potential energy (PE)
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Energy transfer
Rate of energy transfer
Energy associated with
position.
Gravitational PE: mgh
Energy associated with
position in grav. field.
Kinetic energy KE: ½mv2
Energy associated with
motion
Conservative force:
Work done does not depend on path
Non-conservative force:
Work done does depend on path
Mechanical energy ME:
ME=KE+PE
• Conserved if only conservative forces are present
KEi+PEi=KEf+PEf
• Not conserved in the presence of non-conservative forces
(KEi+PEi)-(KEf+PEf)=Wnc
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Work and energy
WORK
POTENTIAL
ENERGY
KINETIC
ENERGY
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Mechanical Energy
Mechanical energy
Gravitational
Potential
Energy (mgh)
Kinetic
Energy
½mv2
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Overview
Newton’s second Law
F=ma
Work
W=(Fcosθ)∆x
Conservation of mechanical
energy
Wnc=0
Closed system
Work-energy Theorem
Wnc=Ef-Ei
Equations of kinematics
X(t)=X(0)+V(0)t+½at2
V(t)=V(0)+at
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Conservation of mechanical energy
A) what is the speed of m1 and m2
when they pass each other?
(PE1+PE2+KE1+KE2)=constant
At time of release:
PE1=m1gh1=5.00*9.81*4.00
PE2=m2gh2=3.00*9.81*0.00
KE1=0.5*m1*v2=0.5*5.00*(0.)2
KE2=0.5*m1*v2=0.5*3.00*(0.)2
Total
At time of passing:
PE1=m1gh1=5.00*9.81*2.00
PE2=m2gh2=3.00*9.81*0.00
KE1=0.5*m1*v2=0.5*5.00*(v)2
KE2=0.5*m1*v2=0.5*3.00*(v)2
Total
196=156.8+4.0v2 so v=3.13 m/s
PHY 231
=196. J
=0.00 J
=0.00 J
=0.00 J
=196. J
=98.0 J
=58.8 J
=2.5v2 J
=1.5v2 J
=156.8+4.0v2
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work
How much work is done by the
gravitational force when the masses
pass each other?
W=F∆x=m1g2.00+m2g(-2.00)=39.2 J
ΣPestart- ΣPepassing=(196.-98.-58.8)=
39.2 J
The work done by Fg is the same as
the change in potential energy
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Friction (non-conservative)
The pulley is not completely frictionless.
The friction force equals 5 N. What is the
speed of the objects when they pass?
(ΣPE+ ΣKE)start-(ΣPE+ΣKE)passing=Wnc
Wnc=Ffriction∆x=5.00*2.00=10.0 J
(196.)-(156.8+ΣKE)=10 J
ΣKE=29.2 J=0.5*(m1+m2)v2=4v2
v=2.7 m/s
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A spring
Fs=-kx
+x
k: spring constant (N/m)
Fs(x=0)=0 N
Fs(x=-a)=ka
Fs=(0+ka)/2=ka/2
Ws=Fs∆x=(ka/2)*(a)=ka2/2
The energy stored in a spring
depends on the location of the
endpoint: elastic potential
energy.
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PINBALL
The ball-launcher spring has a
constant k=120 N/m. A player
pulls the handle 0.05 m. The
mass of the ball is 0.1 kg. What
is the launching speed?
(PEgravity+PEspring+KEball)pull=(PEgravity+PEspring+KEball)launch
mghpull+½kxpull2+½mvpull2
= mghlaunch+½kxlaunch2+½mvlaunch2
0.1*9.81*0+½120(0.05)2+½0.1(0)2=
0.1*9.81*(0.05*sin(10o))+½120*(0)2+½0.1vpull2
0.15=8.5E-03+0.05v2
v=1.7 m/s
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Ball on a track
A
h
end
B
h
end
In which case has the ball the highest velocity at the end?
A) Case A
B) Case B
C) Same speed
In which case does it take the longest time to get to the end?
A) Case A
B) Case B
C) Same time
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Race track
KE PE TME NC
With friction
KE PE TME NC
KE PE TME NC
PHY 231
KE PE TME NC
KE PE TME NC
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A swing
30o
h
L=5m
If relieved from rest, what is
the velocity of the ball at the
lowest point?
(PE+KE)=constant
PErelease=mgh (h=5-5cos(30o))
=6.57m J
KErelease=0
PEbottom=0
KEbottom=½mv2
½mv2=6.57m so v=3.6 m/s
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A running person
While running, a person dissipates about
0.60 J of mechanical energy per step per
kg of body mass. If a 60 kg person develops
a power of 70 Watt during a race, how
fast is she running (1 step=1.5 m long)
What is the force the person exerts on the
road?
W=F∆x P=W/∆t=Fv
Work per step: 0.60 J/kg * 60 kg=36 J
Work during race: 36*(racelength(L)/steplength)=24L
Power= W/∆t=24L/∆t=24vaverage=70 so vaverage=2.9 m/s
F=P/v so F=24 N
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