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PHYSICS 231
Lecture 14: revision
missing ID’s
129020
151313
130786
128820
152180
152183
Remco Zegers
Walk-in hour: Monday 9:15-10:15 am
Helproom BPS 1248
PHY 231
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Chapter 4: Newton’s Laws
 First Law: If the net force exerted on an object
is zero the object continues in its original state of
motion; if it was at rest, it remains at rest. If it
was moving with a certain velocity, it will keep on
moving with the same velocity.
 Second Law: The acceleration of an object is
proportional to the net force acting on it, and
inversely proportional to its mass: F=ma
 If two objects interact, the force exerted by the
first object on the second is equal but opposite in
direction to the force exerted by the second
object on the first: F12=-F21
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Important equations
Newton’s second law: F=ma
Gravitational Force: F=mg
F=Gm1m2/r2
Earth: m1=mearth
r=rearth
g=Gm2/r2=9.8 m/s2
Equilibrium: Fx=0 Fy=0 (object not moving)
FL=0 F//=0
normal force: force perpendicular to surface the object
is resting on, and balancing the component of
the gravitational force perpendicular to the
surface
friction:
F=sn
(s: coef. of static friction)
F=kn
(k: coef. of kinetic friction) s>k
Hooke’s law: Fs=-kx
(k: spring constant)
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example 1
To lift a patient, 4 nurses grip the sheet on which the patient
is lying and lift upward. If each nurse exerts an upward
force of 240 N and the patient has an upward acceleration
of 0.504 m/s2 what is the weight of the patient?
F=ma (only in vertical direction)
4Fnurse-Fg,patient=mpatientapatient
960-mpatientg=0.504mpatient
mpatient=960/(0.504+9.81)=93.1 kg
wpatient=93.1*9.81=913 N
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example 2
A 5.0 kg bucket of water is raised from a well by a rope.
If the upward acceleration of the bucket is 3.0 m/s2, find
the force exerted by the rope on the bucket.
T
F=ma (vertical direction only)
T-mg=ma
T=m(g+a)=5(3.0+9.8)=64 N
Fg
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example 3
A 1000-kg car is pulling a 300 kg trailer. Their acceleration
is 2.15 m/s2. Ignoring friction, find:
a) the net force on the car
b) the net force on the trailer
c) the net force exerted by the trailer on the car
d) the resultant force exerted by the car on the road
Fengine
1000
300
Ftc=-Fct
Fengine=mtotala=1300*2.15=2795 N
Fct=mtrailer*2.15=645 N, so Ftc=-645 N
a) Fcar=2795-645=2150
Fcar=mcar*2.15=2150
b) Ftrailer=Ftc=645 N
2150
c) Ftc=-645 N
d) Ftotal=(21502+(-9800)2=
1E+04 N
PHY 231
mg=1000*9.8
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Ffriction n
A
example 4
1 kg
T
q=20o
Fg
2 kg mass:
1 kg mass:
No sliding:
T
Is there a value for the static
friction of surface A for which
these masses do not slide?
If so, what is it?
0.5 kg
Fg
F=ma (only vertical)
T-mg=ma T-0.5g=2a
F=ma (parallel to the slope)
-Fg//-T+Ffriction=ma
-mgsin(q)-T+smgcos(q)=ma
-3.35-T+9.2s=a
a=0, so T=0.5g=4.9 (from 0.5kg mass equation)
-3.35-4.9+9.2s=0
s=0.9
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PHY 231
example 5
F
A force F (10N) is exerted on
the red block (1 kg). The coef.
of kinetic friction between the
red block and the blue one is 0.2.
If the blue block (10kg)rests on a
frictionless surface, what will
be its acceleration?
Ffriction= kn= kmg=0.2*9.8=1.96 N (to the left)
Fred-blue=-Fblue-red so force on blue block=1.96 N (to the right)
F=ma so 1.96=10a a=0.196 m/s2
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example 6
A rocket is fired from a launching pad. The velocity AND
acceleration of the rocket increase with time even though
the thrust of the engine is constant. Why?
a)The gravitational force becomes smaller because the mass
of the rocket becomes smaller (losing fuel)
b) The gravitational constant g becomes smaller if the rocket
travels away from the earth
c) While traveling away from earth, the gravitational pull
from other planets/sun becomes stronger supporting the
further acceleration
d) Because the density of air becomes small, the friction
becomes less
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Revision: chapter 5
 Work: W=Fcos()x
 Power: P=W/t
 Potential energy (PE)






Energy transfer
Rate of energy transfer
Energy associated with
position.
Gravitational PE: mgh
Energy associated with
position in grav. field.
PE stored in a spring: 1/2kx2 x is the compression of the spring
k is the spring constant
Kinetic energy KE: 1/2mv2 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
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(KEi+PEi)-(KEf+PEf)=Wnc PHY 231
m=1 kg
example 7
A pendulum is pushed with initial
velocity 0.1 m/s from a height of
1 cm. How far does it compress the
spring? (assume m does not rise
significantly after hitting the spring)
1 cm
k=100 N/m
Conservation of ME:
(mgh+1/2mv2+1/2kx2)initial= (mgh+1/2mv2+1/2kx2)final
1*9.8*0.01+0.5*1*0.12+0.=0.+0.+0.5*100*x2 so x=0.045 m
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example 8
A ‘smart’ student decides to save energy by connecting his
exercise treadmill to his laptop battery. If it takes
70 J to move the belt on the treadmill by 1 meter and
50% of the generated energy is stored in the battery, how
‘far’ must the student run to use his 100 W laptop for free
for 2 hours?
Work done by student: W=70*d J
Energy given to the battery 0.5W=35*d J
100 W laptop for 2 hours: 100 J/s*3600*2 s=7.2E+5 J
720 KJ
7.2E+5=35*d so d=(7.2E+5)/35=2.1E+4 m =21km!!!
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example 9
h
450
A block of 1 kg is pushed up a 45o
slope with an initial velocity of 10 m/s.
How high does the block go if:
a) there is no friction
b) if the coefficient of kinetic friction
is 0.5.
A) Conservation of ME: (mgh+1/2mv2)initial= (mgh+1/2mv2)final
0.+0.5*1*102=1*9.8*h+0. So h=5.1 m
B) Energy is lost to friction:
(mgh+1/2mv2)initial= (mgh+1/2mv2)final+Wfriction
[W=Fx=nx=mgcos(45o)h/sin(45o)=0.5*1*9.8*h=4.9h]
0.+0.5*1*102=1*9.8*h+0.+4.9h so h=3.4 m
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example 10
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
example 11
A car (A) can accelerate from 0 to 10 m/s in 5 s. Another
car (B) of the same mass as A can reach 8 m/s in 5 s seconds.
What is the ratio of the power of car A to car B ( PA/PB)?
A)
B)
C)
D)
E)
F)
10/8
8/10
1
10/5
8/5
100/64
P=W/t since t is the same for A and B, P~W.
W=KEf-KEi =0.5m(vf2-vi2)
WA=0.5m102 WB=0.5m82
PA/PB=WA/WB=102/82=100/64
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example 12
A 70-kg diver steps of a 10-m tower and drops, from rest
straight down into the water. If he comes to rest 5.0 m
below the water surface, determine the average resistive
force exerted by the water.
A
(KE +PE )-(KE +PE )=W
i
i
f
f
nc
Choose the final rest point under water
as the zero level.
KEA=0 PEA=mgh=70*9.8*15=10290 J
KEC=0 PEC=mgh=70*9.8*0=0 J
Wfriction=10290 N=(Fcos)x=5F
F=2058 N
Maximum speed (just before hitting the water):
KEA+PEA=KEB+PEB 0+10290=0.5*70*v2+70*9.8*5
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v=14 m/s
PHY 231
B
C