PHYSICS 231
Lecture 39: Review
Remco Zegers
Question hours: Monday 15:00-17:00
Cyclotron building seminar room
PHY 231
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20
chapter 4
100
A person is pushing an ice-sledge of 50kg over a frozen lake
with a force of 100N to the east. A strong wind is pushing
from the south-west and produces a force of20N on the
sledge. a) What is the acceleration of the sledge?
b) if the coefficient of kinetic friction is 0.05, what is the
acceleration?
a) force
Horizontal vertical
person 100
0
wind
20cos(45)
20sin(45)
sum
100+14.1=114.1 14.1
Total force: √(114.12+14.12)=115 N
make sure you
F=ma 115=50a a=2.3 m/s2
b) Ffriction=µn=0.05*m*g=0.05*50*9.8=24.5N understand
ΣF=ma 115-24.5=50a a=1.8 m/s2
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PHY 231
chapter 5
A crate of 50kg is starting to slide from a slope. When it
reaches the bottom, it is caught by a spring with a spring
constant of 1000 N/m. a) If the crate was originally at a height
of 10 m and friction can be ignored, how much is the spring
maximally compressed?
b) if the frictional force is 100N and
the length of the slope is 15m, what is
the maximal compression?
a) use conservation of mechanical E.
at top: Etot=1/2mv2+mgh=mgh=50*9.8*10=4900 J
when spring is maximally compressed:
Etot= 1/2mv2+mgh+1/2kx2=1/2*1000*x2=500x2
4900=500x2 x=3.1 m
b) Wnon-conservative=Ffriction*∆x=100*15=1500N
The mechanical energy just before the block hits the spring:
4900-1500=3400J 500x2=3500 x=2.6 m
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PHY 231
chapter 6
Two objects collide head on. Object 1 (m=5kg) has an
initial velocity of 10m/s and object 2 (m=10 kg) has an initial
velocity of -8 m/s. What is the resulting velocity
a) of the combined object if the collision is fully inelastic?
b) of each of the objects if the collision if fully elastic?
a) inelastic: only conservation of momentum.
m1v1i+m2v2i= (m1+m2)vf
5*10+10*(-8)=15*vf vf=-2m/s
b) elastic: conservation of momentum and kinetic energy
momentum: m1v1i+m2v2i= m1v1f+m2v2f 5*10-10*8=5v1f+10v2f
kinetic energy: (v1i-v2i)=(v2f-v1f)
10-(-8)=v2f-v1f
-30=5v1f+10v2f & 18=-v1f+v2f
v2f=18+v1f
-30=5v1f+10(18+v1f) v1f=-14 v2f=4 m/s
PHY 231
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chapter 10
An ideal gas in a volume of 5m3 at 1 atm is compressed to
a volume of 2m3 and the pressure changes to 1.5 atm. If the
original temperature was 27oC, what is the change in internal
energy of the gas?
original: pV/T=nR 1.0x105*5/(273+27)=1.67x103=nR
p1V1/T1= p2V2/T2 T2=p2V2T1/p1V1=180K
U=3/2nRT ∆U=3/2nR∆T=3/2*1.67x103*(180-300)=-3x105 J
faster: U=3/2nRT=3/2pV
∆U=3/2(p1V1-p2V2)=3/2(1.5x105*2-1x105*5)=-3x105
PHY 231
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Chapter 11
A cube of ice with mass of 1 kg and a base area of 0.01m2
is placed on top of a aluminum plate of 1cm thickness. The
plate is heated from below and the temperature just below
the plate is 500C. Given kAl=238 J/sm0C and
Lice=3.33x105 J/kg, how long does it take to melt all the ice,
assuming no heat is lost.
Q=mL=1*3.33x105=3.33x105 J
P=kA∆T/∆x=238*0.01*50/0.01=1.19x104 J/s
t=Q/P=28 s
PHY 231
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Chapter 11
A 0.2 kg aluminum plate, initially at 200C slides down a 15-m
long surface, inclined at 300 with the horizontal. The force
of kinetic friction exactly balances the component of gravity
down the plane. If 90% of the mechanical energy of the
system is absorbed by the aluminum, what is the temperature
increase at the bottom of the incline? (cAl=900 J/kg0C).
Fg=mg, along the slope: Fg//=mgsinθ=0.2*9.8*0.5=0.98N
Wfroction=F∆x=0.98*15=14.7 J
90% given to the plate: 0.9*14.7=13.2 J
Q=cm∆T ∆T=Q/cm=13.2/(900*0.2) ∆T=0.0730C.
PHY 231
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P(atm)
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Chapter 12.
1
1
2
Consider the process in the figure.
a) How much work is done on the gas?
b) What is the change in internal
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V(m )
energy?
c) how much heat was added to the
system?
a) Work: area under the graph: 1x105+0.5x105=1.5x105J
work done on the gas: -1.5x105 J
b) ∆U=3/2∆(nRT)=3/2∆(pV)=3/2(p2V2-p1V1)=
=3/2(1x105*2-1x105*1)=1.5x105J
c) ∆U=Q+W 1.5x105=Q-1.5x105 Q=3x105 J
PHY 231
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Chapter 12
An engine is operated between a hot and a cold reservoir
with Qhot=400J and Qcold=300J. a) what is the efficiency
of the engine?
The engine is modified and becomes a carnot engine. As a
result the efficiency is doubled. b) what is the ratio
Tcold/Thot. c) what is the maximum efficiency of this engine?
a) efficiency=1-Qcold/Qhot=1-300/400=0.25
b) new efficiency: 0.5=1-Tcold/Thot Tcold/Thot=0.5
c) 0.5 (Carnot engine has maximum efficiency).
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Chapter 12
A block of ice of 1 kg at 00C is melted (L=3.33x103J/kg).
What is the change in entropy?
∆S=Q/T=Lm/T=3.33x103*1/273=12.2 J/K
PHY 231
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Chapter 13
I attach a 2.0 kg block to a spring that obeys Hook’s law and
supply 16J of energy to stretch the spring. After releasing
the block, it oscillates with a period of 0.30 s. What is
the amplitude of the oscillation?
Energy stored in the spring: 1/2kx2=16J
When oscillating, the velocity is zero when the mass is
at maximum amplitude, so 1/2kx2+1/2mv2=1/2kA2=16J
A=√(2*16/k)
ω=√(k/m) ω=2πf=2π/T so T=2π√(m/k)
0.3=2π√(2/k) so k=877 N/m
and thus A= √(32/877)=0.19m
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Chapter 13
Transverse waves travel with a speed of 200 m/s along
a copper wire that has a diametr of 1.5mm. What is the
tension in the wire? (ρcopper=8.93 g/cm3).
v=√(F/µ) so F=v2 µ
v=200m/s
µ=mass of wire per unit length (1 meter)
=density*volume: 8.93x103kg/m3*(πr2*1)
µ=8.93x103*1.77x10-6=1.58x10-2 kg/m (diameter-> radius)
F=632N
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Chapter 14
A football fan upset that his team is losing tosses out
his battery powered radio out of a 20m high window.
What is the frequency of the sound from the radio just
before it hits the ground, relative to the frequency when
it just drops out the window (assume: initial velocity 0).
f’=f(v+v0)/(v-vs)
v: velocity of sound (343 m/s)
v0: velocity observer (0 m/s)
vs: velocity source.
vs(t)=v0+at=0-9.8t
x(t)=20-0.5gt2 0=20-0.5*9.8*t2
t2=40/9.8 t=2s so v=-19.6 s
radio is moving away from the observer, so negative sign.
f’/f=v/(v-vs)=343/(343-(-19.6))=0.95
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PHY 231
Chapter 14
If the distance between a point sound source and a dB
detector is increased by a factor of 4, what will be the
reduction in intensity level?
Intensity~1/(distance from source)2
I~1/r2
I2/I1=r12/r22=1/42=1/16
β=10log(I/I0) I0=1x10-12 W/m2
β2-β1=10[log(I2/I0)-log(I1/I0)]=10log(I2/I1)=10log(1/16)
β2-β1=-12dB
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Chapter 8
An 800-N billboard worker stands on a 4.0-m scaffold
supported by vertical ropes at each end. If the scaffold
weighs 500N and the worker stands 1.0m from one end,
what is the tension in the ropes?
L
R
1.0m
Net force in vertical direction must be 0:
FL+FR-wman-wscaffold=0 FL+FR-800-500=FL+FR-1300=0
Net torque must be zero. Choose 0 at ‘R’.
FL*4-800*1-500*2=4FL-1800=0 FL=1800/4=450N
so FR=1300-FL=850N.
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Chapter 7
A roller coaster, loaded with passengers, has a mass of
mass of 2000 kg. The radius of the curvature of the track
at the bottom point of the dip is 24m. If the vehicle has a
speed of 18 m/s at this point, what force is exerted on the
vehicle by the track?
two forces: gravitational force and centripetal force.
gravitational force: mg=2000*9.8=19600N
centripetal force: mv2/r=2000*182/24=27000N
Summed: 46600N.
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