CONCEPTUAL rrgs/C
Name Date
PRACTICE PAGE
Chapter 2 Newton's First Law of Motion-Inertia
Static Equilibrium
1. Little Nellie Newton
wishes to be a
gymnast and hangs
from a variety of
positions as shown.
Since she is not
accelerating. the net
lorce on her is zero.
That is, '£F = O.This
means the upward
puu 01the rope(s)
equals the downward
pUll of gravity.
She weighs 300 N.
Show the scale
readlng(s) for each
case.
bOON .••
400 N
2. When Burl the painter stands In the
exact middle 01 his staging. the left
scale reads 600 N. Fill in the reading
on the right scale. The total weight
of Burl and staging must be
12!lll N.
•. N
3. Burl stands larther from the left. Fill
In the reading on the right scale.
•• 1200 N
<I In !\ smYImood. BUllIclanglli<' In,,,, I;the rigl!ll end. Fill inth!! rQadlng onJ'
right scale. '
..sl...;tt
~..Jt!
3
•
CONCEPTUAL PRACTICE PAGE
Chapter 2 Newton's First Law of Motion-Inertia
The Equilibrium Rule: IF =0
1. Manuel weighs 1000 N and stands In the
middle of a board that weighs 200 N. The
ends 01the board rest on bathroom scales.
(We can assume the weight of the board
acts at its center.) Fill in the correct weight
reading on each scale.
850 N '<.00 N
1000 N
2. When Manuel moves 10the left as shown.
the scale closest to him reads 850 N. Fill
in the weight for the far scale.
200N
/000 N
13 TONS
3. A tz-ton truck is one-quarter
the way across a bridge that
weighs 20 tans. A 13-ton lorce
supports the right side of the
bridge as shown. How much
support force is on the left side?
20 TONS
NQI'!OOI: _ .1~ •• N
4. A 1OOQ-N crate resting on a
surface is connected to a
500-N block through a frictionless
pulley as shown. Friction between
the crate and surface is enough
to keep the system at rest. The
arrows show the forces that act
on the crate aM th" biecl<.Fin Il'l
the magnitude ot each lorce.
Tension '. _ ~l!O__ N
crate
w= __tOOILN
ll'OO
1
!U~bk .~ L¥S
W·" _~~~_N
5 il the crate and block In the preceding question move at constant speed. the tension In the rope
Gthe saRiiD Uncreases) [decreases] .
The sliding system ISthen m [StatICeqUilibrium) amic equilibnu J..••llt I
~ ••if'.
4
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CONCEPTUAL I: PRACTICE PAGE
Chapter 2 Newton's First Law of Motlon-lnt.rt~~..M.,5r'
VlICton and Equilibrium (...., ~~ ~_~1(.c.
n ~1, Nellie Newton dangles from a
vertical rope in equilibrium:
:l:F = 0, The tension In the rope
(upward vector) has the same
magnitude as the downward
pull of gravity (downward vector).
2, Nellie is supported
by two vertical ropes.
Draw tension vectors
to scale along the
direction of each rope.
3, This time the vertical ropes have
different lengths, Draw tension
vectors to scale for each of the
two ropes,
4. Nellle is supported by three
vertical ropes that are equally
taut but have different lengths.
Again, draw tension vectors to
scale for each of the three ropes.
Rope te:ItS'on does depend on the: angle the: rope m<lk~ WIth
the vertical, 0$ Practice Pages far Chapter <> Willshow!
)' )
Name Date
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CONCEPTUAL "'slc PRACTICE PAGE
Chapter 3 Linear Motion
Free Fall Speed
1. Aunt Minnie gives you $10 per second for 4 seconds.
How much money do you have after 4 seconds?
2, A bali dropped from rest picks up speed at 10 m/s per second.
After It falls for 4 seconds, how fast is it going?
3, You have $20, and Uncie Harry gives you $10 each second for 3 seconds,
How much money do you have after 3 seconds?
4, A ball is thrown straight down with an Initial speed of 20 m/s,
After 3 seconds, how fast is il going?
5. You have $50, and you pay Aunt Minnie $10/second.
When will your money nun out?
6, You shoot an arrow straight up at 50 rn/s.
When will it run out of speed?
,',
, '
1
7, So what will be Ihe arrow's speed 5 seconds after you shoot it?
8. What will its speed be 6 seconds after you shoot it?
Speed after 7 seconds?
Free Fall Distance
1. Speed is one thing; distance is another, How high is the arrow
when you Shoot up at 50 mts when rt runs out of speed?
2. How high will the arrow be 7 seconds after being shol up at 50 m/s?
b. What is the penny's average speed during its
3-second drop?
c. How far down is the water surface?
7
CONCEPTUAL ,.
Chapter 3 Linear Motion
Accefetation of Free Fait
•
PRACTICE PAGE
A rock dropped from th!!>top of a cliff
picks up speed as iltal/s. Pretend
that a speedomoter and odometer
are attached to the rock to indicate
readings of speed and distance at
t-ssconc intervals. Both speed and
distance are zero at lime; zero (see
. sketch). Note that after faUltlg 1 second,
the speed reading is 10 'm/s and the
distance fallen is 5 m. The reedings
of succeeding seconds of faU are not
shown and are left for you 10 complete.
So draw the position of the speedometer
pointer and write in the correct ocometer
reading for eacn time. Use g = 10 m/s'
and neglect air resistance.
YOU NUO TO KNOW:
IMtaotoMOllS 'fle.~d of foil
h'Qrt'l re.t:
v » gt
[)is~ont~ failefl from rest:
a » V,"<09< t
or
o> '12 f
1. The speedometer reading increases the same
amount, --1.1L- mls, each second.
This increase in speed per second is called
Z. The distance lallen increases SI' thE; square of
tne ~.
3 It ,I takes 7 seconds te reach the gf~\und,
then ns spet1d atlirlpact is -.-ZL_ m/s,
the total distance fallen is ~_.. m,
and rtsaccllleration of fall just beJor"" impact is
-1!L_ m/s".
I
)
{
t 0
I
I
I
t.// ,
f
r
cl>
I.
<)
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Name Date
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CONCEPTUAL rtr-Cllapt~r 3 Linear Motion
Hang17me
I'\'tACTIC!: PAGE
Some athtetlilS and dancers have great jumping abWty. When leaping. they seem to
momenlanly "hang in the air" and defy gravity. The lime that a jumper is airborne with feet off the
grO\lnd is called hang time. Ask your friends ID estimate tne hang time of the great jumpers. They
may say two or three seconds. But surprisingly, the hang time of the greatest jumpers is most
always less than 1 second! A longer time is one of many illusions we have about nature.
To better understand this, find the answers to the following questions:
1. H YO~alSlfepoff ,:tt"ble anhd ~ tak
l
es Spud of free fall e ~c«kraticn x time
one·" secnn" f) reac th~foor, '. J. . f. d.
what Will be the speed when you ' 10 ",Is x n~mb e r 0, seccn as
meet the floor? , lOt m.
v = gt ee 10 mJs' x t " 5 mJs
2. What Will be your averag~ sp~ed of fall?
v ~ 2._+5 mJ~ " 2.5 mls
2
---_.,-Distance' average speed x time.
3 What will be. the distance of fall?
1
d '" vI~' 2.5 mls < 2s " 1.25 1Il
4. So how high. is th~ surtace of the ta,bl" above th" floor?
Jumping ability is best measured by a standing' vertical jump. Stand facing a watt
with j"et flat on th" floor and- arms extended:·upward. M<l.l<ea m"rt< on Ihe wall at
the top of your re<lch. Then make your jump and at the peak make another mark.
The distance between thes~ IWO m"r.ks measures your verticallEiap. If It's more
than 0.$ meters (2 feet), you're exceptional
5. What is your vertical jumping distance? _111A!'!:.!~..§1
e. Caiculate your personal hang time using the formula d ~ 1f.? gl' (Remember that
hang time is the time that you move upward + the time you return downwar.d.)
Aim~sta.ybodycah sof.ly step eff Q 1,2~'1l\(4-filet) hi~htcble,
Con. a"ybody In YQurs,chQolJtll'\'\P from the floor up ".toth" $Q"". t.o..bl.? ~'. .' ••'
~ '. '. ~ "{~
S~,,)
There'$ a big difterence in.ho" high you conre4ch Md ho"", high yo" rcise '\
~
tour' c<:nf",rof lWGi'fity' \'Inen you Ju",p. cwn. bCi$.k~liJ<)!1srer Michad
-;:;, JordCin in h,s 1'1'1",. couldn't qUl'e ro,se hIS body 1.25 metE;rs hIgh, olthough
'it he could eoslly reoth higher than fhe l1tore-lhon.3-meter high bcsker.
Here we're talking about vertical motlOh HOW abo u!fUl1l11nglumps? We'l/ see In. Chapter 10 that the
height 01 ajump depends oniy on Ille IOhlpsr's vmtlcal $p$ed allaunCh While arrborne the lumper s
honzonlal speed remains eonstanr while lhe vertical speed undergoes acceleration due to gravity.
While airbome, no amount of Jeg or afm pumping or other bodily motions can cflange lI,?ur flang time.
~!<t:
)
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9
•CONCEPTUAL PRACTICE PAGE
Chapter 3 Linear Motion
Non-Accelerated Motton
1. The sketch shows a ball rolling at constant velocity along a level floor. The ball rolls from the
first position shown to the second in 1 second. The two positions are 1 meter apart. Sketch
the ball at successive 1-second intervals all the way to the wall (neglect resistance).
o
a. Did you draw successive ball positions evenly spaced, farther apart, or closer together? Why?
EVENLY SPACED-EQUAL DISTANCE IN EQUAL TIME -> CONSTANT y
b. The ball reaches the wall with a speed of _1_ mls and takes a time of .....L seconds.
2. Table I shows data of sprinting TABLE I
speeds of some animals.
Make whatever computations
necessary to complete the table.
ANIMAL OISTANCE 'TIME SPEED
CHEETAH 751'l'1 3s 2S M/s
GREYHOUND 160 m 10 s " IIVs
6AZElLE I I<m 0.01 h IOOkmA>
'TURTLE 3o,,,, 305 1 em/s
Accelerated Motion
3. An object starting from rest gains a speed v = at when it undergoes uniforrn acceleration.The
distance it covers is cl = 1/2 af. Uniform acceleration occurs for a ball rolling down an inclined
.plane. The plane below is tilted so a ball picks up a speed of 2 mis each second: then its
acceleration a = 2 m/s'. The positions of the ball are shown at t-seoond intervals. Complete
the six blank spaces for distance covered and the four blank spaces for speeds.
a. Do you see thatlhe total distance from the starting point increases as the square of the time?
This was discovered by Galileo. Ifthe incline were to continue, predict the bali's distance from
the starting point for the next 3 seconds.
YES; DISTANCE INCREASES AS SQUARE OF TIME' 36 rn 49 rn. 64 m.
b. Note the increase of distance between ball positions with time. Do you see an odd-integer
pattern (also discovered by Galileo) for this increase? If the incline were la continue, predict
the successive distances between ball positions for the next 3 seconds.
YES; 11 m, 13 rn, 15 m.
10
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Name Date
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CONCEPTUAL "",Ic: PRACTICE PAGE
Chapter 4 Newton's second Law of Motion
~~~t ~~.
Learning physics is learning the connections amo[1Qconcepts in nature, and ~f~also learning la distinguish between closely-related concepts. Velocity and~· ..
acceleration, previously treated, are often confused. Similarly in this chapter, ..
we find that mass and weight are often confused. They aren't the same!.(':
Please review the distinction between mass and weight in your textbook, .. "': ..
To reinforce your understanding of this distinction, circle the correct answers below:
Comparing the concepts of mass and weight, one is basiC-fundamental-depending only on the
internal makeup of an object and the number and kind of atoms that compose ~.The concept that
is fundamental is e [weight].
The concept that add~ionaflydepends on localion in a gravitational field is [mass] ~
e[Weight] is a measure of the amount of matter in an object and only depends on the
n er and kind of atoms thet compose it.
It can correctly be said that ~ [weight] is a measure of "laziness' of an object.
[Mass] ~ is related to the gravitational force acting on the object.
[Mass] ~ depends on an object's location, whereas ~ [weight] does not
In other words, a stone would have the same W [We~ether it is on the surface of
Earth or on the surface of the Moon. However, ~s mass] ~ depends on its tocanon
On the Moon's surface, where gravity is only about 116" Earth gravity ~ [weight]
[both the mass and the weight] 01the stone would be Ihe same as on .
While mass and weight are not the same, they are directly proportlona [inversely proportional]
to each other. In the same location, twice the mass has the weight.
The Standard Internation I (SI) unit of mass is the @~'3> [newton), and the SI unit ot torce
is the (kilogram] newton
In the Un~edStales, It is common to measure the mass of something C;uring its gravitational
pUlito Earth, ~sweight. The common unit of weight in the U.S. Isthe pound [kilogram] [neWlon].
Support
Force
When I step on a weighing scale. two forces act on it; a downward
pull of gravity, and an upward support force. These equal and
opposite forces effectively compress a spring inside the scale that
is calibroted to show weight, When in equilibrium. my weight" mg.
thorlx to Daniela Taylor
11
•
CONCEPTUAL PRACTICE PAGE
Chapter 4 Newton's Second law of Motion
Converting Mass to Weight
Objects w~h mass also have weight (although they can be weightless under special conditions). If
you know the mass of something in kilograms and want ~s weight in newtons, at Earth's surface,
you can take advantage 01 the formula that relates weight and mass.
Weight = mass x acceleration due to gravity
W=mg
This is in accord with Newton's 2nd law, written as F = ms. When the force of gravity is the only
force, the acceleration of any object of mass m will be g, the acceleration of free fall. Importantly,
9 acts as a proportionalily constant, 9.8 Nl1<g, which is equivalent to 9.8 mls'
Sample Question: from F = ma, we see tMt the unit of
How much does a I-kg bag of nails weigh on Earth? force equols the units [k9 x m/s']. Can
~
YOu see the units (m/s'] 0 (N/kg]?
W = mg = (1 kg}(9.8 m/s') = 9.8 m/s' = 9.8 N. J
or simply. W = mg = (1 kg)(9.8 Nlkg) = 9.8 N. ':.:;r
Answer the following questions:
Felicia the ballet dancer has a mass of 45.0 kg.
441 N
2. Given that 1 kilogram of mass corresponds to 2.2 pounds at Earth's surface, ", LB
what is Fellcia's weight in pounds on Earth? ...
--~.~~
111..5 N
1. What is Felicia's weight in newtons at Earth's surface?
3. What would be Felicia's mass on the surface of Jupiter?
4. What would be Falicia's weight on Juplier's surface,
where the acceleration due to gravity is 25.0 m/s2
?
Different masses are hung on a spring scale calibrated in newtons.
The force exerted by gravily on 1 kg = 9.8 N.
5. The force exerted by gravity on 5 kg = ~ N.
6. The force exerted by gravity on ........1.lL- kg = 98 N.
Make up your own mass and show the corresponding weight:
The force exerted bygravilyon _,,-- kg = _._ N.
'ANY VALUE FOR kg AS LONG AS THE SAME VALUE IS MULTIPUED
8Y9.8FOR N.
By whatever means (spring scales,
measuring balances, etc.), find the
mass of your physics book. Then
complete the table.
OBJECT MASS WEIGHT
MELON I kg 'loa to!
APPLE 0.1 ~. iN
BOOK
A FRIEND 60 k9 518 N
...L·t"'I
~Wll.
12
Name Date _
CONCEPTUAL '1tJIsk
Chapter 4 Newton's Second Law of Motion
A Day at the Races with a =Fhn
PRACTICE PAGE
In each situation below, Cart A has a mass of 1 kg. Circle the correct answer (A, B, or Same for both).
1. Cart A is puned with a force of 1 N.
Cart B also has a mass of 1 kg and is
pulled with a force of 2 N
Which undergoes the greater acceleration?
lA] ® [Same tor both]
A~~~
3. Cart A is pulled w~h a force of 1 N.
Cart B has a mass of 2 kg and is pulled
with a force of 2 N.
Which undergoes the greater acceleration?
[AJ [BI <i&ame for bolD
A~~~
B~i1t:;-
5. This time Cart A is pulled with a force of 4 N.
Cart 8 has a mass of 4 kg and is pulled with
a force 014 N
Which undergoes the grealer acceleration?
@ {8] (Same tor both]
/
-'
2, Cart A is pulled With a torce of 1 N
Cart B has a mass of 2 kg and is also
pulled with a force of 1 N,
Which undergoes the greater acceleration?
@ [B] [Same for both]
A~~~
B~Ul~t~
4. Cart A is pulled with a force of 1 N.
Cart B has a mass of 3 kg and is puued
with a force of 3 N.
Which undergoes the greater acceleration?
[A] [B] €me for~
A~ :!~I~t)@+
~~l!::t~
6. Cart A is pulled with a force of 2 N.
Cart B has a mass of 4 kg and is pulled
with a force of 3 N.
Which undergoes the greater acceleration?
®[B] [Same for both]
A~3!~2Lx:--
B~~;>o--
'tbcnx to ()earl Boird
)
•CONCEPTUAL PRACTICE PAGE
Chapter 4 Newton's Second Law of Motion
Dropping Muses and AccelerBting Cart
1. Consider a 1-kg cart being pulled by a 10-N applied lorce
According to Newton's 2nd law. acceleration 01 the cart is
El=
F
m
~
lit tON
2
Tkg" = 10m/s .
Thi&is the Same as the ncceleretlen of free fall, rbecouse a
force cqU41to the C4rt'.s weight accelerates it.
2. Consider the acceleration 01 the cart when the applied lorce
is due to a 1D-N iron weight attached to a string draped over
a pulley. Will the cart accelerate as before, al1 0 m/s'?
The answer Is no, because the mass being accelerated is
the mass of the cart plus the mass of the piece of iron that
pulls it. Both masses accelerate. The mass of the 1D-N
iron weight is 1 kg-so the total mass being accelerated
(cart + iron) is 2 kg. Then,
The pulley changes only
the direc:tion of the force.
a=
F
m
10N 2
2kg = Sm/s .
Don't forget; the tota!lJIQSS' of a system
includes the mass of the hanging irol\.
Note this is half the ecceleretten due to gravity alone. g. So
the. acceler<ltlon of 2 kg produe&d by the weight of 1 kg ill g/2.
a. Find the acceleration of the 1-kg cart When two
identical 1D-N weights are attached to the string
a"
F
m
.sa.:».applied force
total mElSS
'2.0N
'3 I<j
Here we simplify and say
g: 10 m/s2
.
14
•
CONCEPTUAL PRACTICE PAGE
Chapter 4 Newton's Second Law of Motion
Dropping Masses and Accelerating Cart-continued
b. Find the acceleration of the 1-kg cart when the
three identical 1D-N weights are attach to the
string.
a"
F
m
30N
4 Kt
applied force
total mass
c. Find the acceleration of the 1-kg cart when four identical10-N weights (not shown)
are attached to the string.
a"
F
m
applied force
total mass
d. This lime 1 kg of iron is added to the cart, and
only one iron piece dangles from tne pulley.
Find the acceleration of the cart.
F
ION------_._'3
~I
a"
applied force
total massm
_a__.__ mis'.
~,.:3__ m/s'.
The force due to gravity on Q mass m is mg_
So grovltotloncil force on 1 kg Is (1 kgXl0 m/SZ) " 10 N.
e. Find the acceleration of the cart when it carries
two pieces of iron and only one iron piece dangles
from the pulley.
a=
F
m
ION
4 1<,.
applied force
total mass
15
w
00
CONCEPTUAL PRACTICE PAGE
Chapter 4 Newton's second Law of Motion
Dropping Masses and Accelerating Cart-continued
f. Find the acceleration Ofthe cart when it carries
3 pieces of iron and only one iron piece dangles
from the pulley.
F applied force
lolal mass
a=
m
g. Find the acceleration of lhe cart when it carries
3 pieces of iron and 4 pieces of iron dangle from
the pulley.
a=
F
m
applied 10rce
total mass
______ m/s".
___ 5 m/sa
h. Draw your own combination of masses and
find the acceleration.
a=
F
m
applied force
total mass
1(,
/
______ m/s'.
J.••ilt••.
1
~~4'.
Name Date __ ----
CONCEPTUAL WC: PRACTICE PAGE
Chaptar 4 Newton's second Law of Motion
Force and Aceslsmtlon
1. Skelly the skater, total mass 25 kg, is propelled by rocket power.
a. Complete Table I (neglect resistance).
TABLE I
r:ORCE I>.CCELERATION
100 N 4m1s
200N am/a
2.50N 10 m/sa
F
a =
26mg
b. Complete Table 1I10r a constant 5O-N resistance.
TABLE 11
r:ORCE ACCELERATION
50 N o m/so
100 N ?m/ ••
200N 6m's
F·50N
a = 26 kg
2. Block A on a horizontal friction-free lable is accelerated by a force from IIstring attached
to Block B of the same mass. Block B falls vertically and 'I
drags Block A horizontally. (Negiecllhe string's mass). A
CIrcle the correct anSWets: ~
a. The mass at the system (A + B) is [ml ~
b. The force that accelerates (A + B) is the weight of lA) ® lA + B].
c. The weight of B is Img~2 mg].
d. Acceleration of (A + B) i ~ ~l (more than g].
e. Use a = ~ to show the acceleration of (A + B) as a fraction of g. a = ;:
If 8 WE\'e allowed to foil by itself, not dro99in9 A,
then WOJk:tl'tits QCl;elErotionbe q?
2
Yes, beceese the force ~l oc.c.elerotes
i1 would only be QC.ting on its own
moss - 1101 twice the moss!
17
/
CONCEPTUAL PRACTICE PAGE
Chapter 4 Newton's second Law of Motion
Force and Accelartition-continued
3. Suppose Block A is still a 1-kg block, but B is a low-mass feather (or a coin).
a. Compared to the acceieration of the system of 2 equal-mass biocks
the acceleration of (A + B) is ~ [more]
and is~se to z~ [close to g).
b. In this case. the acceleration of B is
[practically that of freefallJ ~
A
4. Suppose A Is the feather or coin. and Block B has a mass of 1 kg.
a. The acceleration of (A + B) here is [close to zero] ~
b. In this case, the acceleration of Block B is
[practically that of free fall [nearly zero].
5. Summarizing We see that when the weight of one object causes the acceleration
~the range of possible accelerations is between
zero and g zero and infinity] [g and infinity).
6. For a change of pace, consider a ball that rolls down a uniform-slope ramp.
a. Speed of the ball is [decreasing] [constant] Eeas",
b. Acceleration is [decreasing] ~[increaSing].
c. If the ramp were steeper, acceleration would be<;;;;j)[the same] (less].
d. When the ball reaches the bottom and rolls along the smooth level surface, it
'-"r:;:::18
B
Name Date
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CONCEPTUAL fQIs/l PRACTICE PAGE
Chapter 4 Newton's second Law of Motion
Friction
N
w
~
f ~W
N
11 ~
P
l
f W
"1/
1. A crate filled with delicious Junk food rests on a horizontal floor.
Only gravity and the support force of the floor act on it, as shown
by the vectors tor weight W and normal force N.
a. The net force on the crate is ~ [greater than zero].
b. Evidence for this is NO ACCELERATION
2. A slight pull P is exerted on the crate, not enough to move i!
A force of friction f now acts,
a. which is [less than)~ [greater than] P.
b. Net force on the crate is ~ [greater than zero].
3. Pull P is increased until the crate begins to move. It is pulled SO
that tt moves With constant velocity across the floor.
a. Friction fis [less than] ~ [greater than) P.
b. Constant velocity means acceleration is ~more than zero].
c. Net force on the crate is [less than] <l.§Ual toDmore than] zero,
4. Pull P is further increased and Is now greater than friction t.
a. Nef force on the crate is [less than] [equal to] ~
zero.
b. The net force acts ~he right, so acceleration acts
toward the [left] ~
5. 11the pulling force P is 150 N and the crate doesn't move,
what is the magnitude of f? -l§lt!!!
6. If the pulling force P is 200 N and the crate doesn't move, what is the magnitude of f? .........200...
7. 11the force of sliding friction is 250 N. what force is necessary to keep the crate sliding
at constant velocity? ~
when the pulling force is 250 N?
6. lithe mass of the crate is 50 kg and sliding friction is 250 N, what is the acceleration of the crate
300 N?o mls' 1 mls' 500 N? . .--Ulls'
.19
CONCEPTUAL PRACTICE PAGE
Chapter 4 Newton's second Law of Motion~
Failing and Air Resistance '
R'O
Bronco skydives and parachutes from a stationary Q.
helicopter. Various stages of fall are shown in
positions a through t. Using Newton's 2"" law. W ' 1000 N
b ~'4ooN
-lW'1QOON
~'I000N
c -,w'I000N
F W. Ra=.....!J!!L=---
m m
find Bronco's acceleration at each position
(answer in the blanks to the right). You need to know
that Bronco's mass m is 100 kg so his weight is a
constant 1000 N. Air resistance R varies with speed
and cross-sectional area as shown.
Circle the correct answers:
1 When Bronco's speed is least, his acceleration is
[least] ~
2. In which position(s) does Bronco experience a
downward acceleration?
~ICl [d] [e] If]
3. In which posltlon(s) does Bronco experience a
upward acceleration?
raj [C]~[f][b]
4. When Bronco experiences an upward acceleration,
his velocity is ~ [upward also].
5. In Which posrtlon(s) is Bronco's velocity constant?
[a] [b] Gi) [d] [e] G1i)
6. In which position(s) does Bronoo experience terminal
velocity?
[a] rb] (ii) [d] [e}G)
7. In which position(s) is terminal velocity greatest?
[a] [bl <.;:>[dl le} [f]
8. If Bronco were heavier, his terminal velocity would be
<!iir~teD['eSSJ (the same].
20
/
R'I000N
W'1000N
a = =:1Jl..D:!lE
a = ..!!JIJl£
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CONCEPTUAL rtJ PRACTIce PAGE
Chapter 5 Newton's Third Law of Motion
Act/on and Reaction Pairs
1. In the example below, the actlon-reaotlon pair is shown by the arrows (vectors), and the actionreaction
described in words. In a through g, draw the other arrow (vector) and state the reaction
~"-" ;M"·~"-II'~"~'J~'"f~,j~~ ~-+ " - .'):) "'-J_
-r;/I.
Fist hits wall. Head bumps ball. Windshield hits bug.
b. BUG HITS WINDSHIELD.
~
Wall hits fist. a. BALL BUMPS HEAD.
4;f/:- -7/1
Bat hits ball.
c. BALL HITS BAL
Hand touches nose. Hand pulls flower.
e. FLOWER PULLS ON
HAt&.
STUDENT DRAWING
(OPEN)
Athlete pushes bar upward. h. THING A ACTS ON
I!i!J'iG.B.
ATHLETE DOWNWARD
g. BALLOON SURFACE ~
COMPRESSED AIR UPWARD.
IHING B ACTS ON
2. Draw arrows to show the chain of at least six parts of action-reaction forces below.
21
I
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CONCEPTUAL ft.,. PRACTICE PAGE
Chapter 5 Newton's Third law of Motion
Interactions
1. Nellie Newton holds an apple weighing 1 newton at rest on Ihe palm of her hand. The force
vectors shown are the forces that act on the apple. ~ __. N.
a. To say the weight of the apple is 1 N is to say that a
downw~onal force of 1 N is exerted on the
apple b [her hand).
b. Nellie's hand supports the apple.Withnormal force N,
Whichacts in a direction 0 osite to W. We can say N
[equals W] Sthe same magnitude as Wj. . ,
c. Since the apple is at rest, the net force on the apple I~nonzeroJ.
d. Since N is equal and opposite to W. we [can] ~saY that Nand W
comprise an action-reaction pair. The r:son is because action and reaction always
[act on the s:me Ob=-~ ~jffeLt o~ie~nd here we see Nand W
q;;;ctjnoc@ the ~ [acting on different Objects].
e. In accord with the rule, "If ACTION is A acting on S, then REACTION Is B acting on A,"
if We sa acti is Earth pulling down on the apple, then reaction;s
[the apple pulling up on Earth [N, Nellie's hand pushing up on the apple].
f. To repeat for emphasis. we see that Nand Ware equal and opposite to each other
[and comprise an action-reaction pair] 1Iitdo not comprise an action-reaction paid:>
To identify a pair of o.ction-mlclion Ioeces in OI>{ Siluoli"'l first identify
the poir of inleroct i"9 cbjects irnoNed. Something is interocting wi1h
something ,Ise: In this CQSethe vmole E<lrlh is inlerocling (~i\atiorollv)
wi1h the llpple. So Enrlh pt;lIs OOwllWord0I11hc apple (mUll ocfun).
-..I\il< the apple pulls upward on Earih ("""fun)
[slilllwice the magnitude of W]. and
[still W· N, a negative force}.
22
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CONCEPTUALtAysic PRACTice PAGE
Chapter 5 Newton's Third law of Motion
vectors and the Parallelogram Rule
1. Whentwo vectors A and B are at an angle to each other, they add to produce the resultant C
by the parallelogram rule. Note tnat C is the diagonal of a parallelogram where A <lnd Bare
adjacent sides. Resultant C is shown In the· first two diagrams, q and b. Construct resultant C in
diagrams c and d. Note that in diagram dyou form a rectangle (a special case of a parallelogram).
r--/i
~
Q d B
2. Below we see a top view 01an auplane being blown oft course by wind in variou irections.
Usethe parallelogram rule to show the re. g speed and direction of travelloi ~ case.
In which case does the airpiane tr';V$lIraste fcross the ground: ,,'_...9 ... (jlo esn y
3. To the right we see the top views of 3 motorboats crossing
a river. All have the same speed relative to the water, and
ail experience the same water flow.
Construct resultant vectors showing the speed and
direction of the boats.
a. Which boattakes.me shortest path to the opposite shore"
b. Which boat reaches the opposite shore lirst?
__ b
c. Which boat provides the fastest ride?
_..-J;.
2'\
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CONCEPTUAL PRACTICE PAGE
Chapter 5 Newton's Third Law of Motion
Velocity Vectors and Components
Draw the resultants 01 the lour sets of vectors below.
\
\
2. Draw the horizontal and vertical components 01 the lour vectors below.
3 She tosses the ball along the dashed path. The velocity vector, completa with its
horizontal and vertical components, is shown at position A. Carefully sketch the
components for positions Band C,
a, Since there is no acceleration in the horizontal direction, how does the horizontal component
01 velocity compare tor positions A, B, and C?
b. What Is the value 01 the vertical component 01 velocity at position B?
c. How does the vertical component of velocity at position C compare with that 01 position A?
_ EQUAL AND OPPOSITE
)
Name Date
CONCEPTUAL WIJ· PRACTICE PAGE
Chapter 5 Newton's Third Law of Motion
Force and Velocity Vectors
1. Draw sample vectors to represent the terce
of gravity on the ball in the posinons shown
below after it leaves the thrower's hand.
(Neglect air resistance.)
2. Draw sample bold vectors to represent the
velocity 01 the ball in the positions shown below.
With lighter vectors, show the horizontal and
vertical components 01 velocity lor each position.
"
,/1';" 1""1\\, t:"~,
.: \ ~: ~\
.:1 f\:, K
L~_-i..~ ,3 a Which velOCity component m the prev+us queston remains constant? Why?
HORIZONTAL COMPONENT CONSTANT -NO HORIZONTAL FORCE ON BALL
b. Which velocity component changes along the path? Why?
T
4. It is important to distinguish between force and velocity vectors. Force vectors combine with
other force vectors, and velocity vectors combine with other veiocity vectors. Do velocity
vectors combine with force vectors?
5. All forces on the bowling ball, weight (down) and support of alley (up), are shown by vectors
at its center before it strikes the pin a.
~~;i~ectors 01 all the lorces that act~on the ball b when it strikes the pin, and c after itr!strike~,
a b ~-== I\} =I,.
. ': .:
. ,
~.: ." ~
thanx to HOwle 9rorId
2.<;
!/ )
•
)
Chapter 5 Newton's Third Law of Motion
Force Vectors and the Paralfelogram Rule
1. The heavy ball is supported in each case by two strands of rope. The tension in each stral'd
is shown by the vectors. Use the paralielogram rule to find the resultant of each vector pair.
Name Date
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CONCEPTUAL fIvs PRACTICE PAGE
Chapter 5 Newton's Third Law of Motion
Force-Vector Diagrams
CONCEPTUAL PRACTICE PAGE
~--a. is yourresuitant
vector the same for
each case?
--.Y§
b. How do you think the resultant vector compares
with the weight of the ball?
SAME (BUT OPPOSITE DIRECTION)
2. Now let's do the opposite of what we've done above. More often, we know the weight of the suspended
object, but we don't know the rope tensions. In each case below, the weight of the ball is shown
by the vector W. Each dashed vector represents the resu~ant of the pair of rope tensions. Note that
each is equal and opposite to vectors W (they must be; otherwise the bell wouldnl be at rest).
a. Construct parallelograms where the ropes define adjacent sides
end the dashed vectors are the diagonais.
b. How do the relative lengths of the sides of
each parallelogram compare to rope tension?
c. Draw rope-tension vectors, clearly showing
their relative magnitudes.
- - .- • A'
I
I
I
w ww w
3. A lantern is suspended as shown. Draw vectors to show the relative
tensions in ropes A, B, and C. Do you see a relationship between your
vectors A + B and vector C? Between vectors A + C and vector 87
YES:A+B= -C A+B= -13
?fi
In each case, a rock is acted on by one or more forces. Using a pencil and a ruier, draw an accurate
vector diagram showing all forces acting on the rock, and no other forces. The first two cases are
done as examples. The parallelogram rule in case 2 shows that the vector sum of A + B Is equal and
opposite to W (that is, A + B = oWl. Do the same for cases 3 and 4, Draw and label vectors for the
weight and normal support forces in cases 5 to f 0, and lor the appropriate forces in cases f 1 and 12.
B
7. Decelerating due to friction
N
10. Static 11. Rock in lrea.fall
w
27
3. Static
w
6. Sliding at constant speed
without friction
N
g. Rock slides
(No friction)
w
w
to imCoUM -JI..,i1t.,./
'1ilI,J1t.
CONCEPTUAL PRACTICE PAGE
Name Date
Chapter 5 Newton's Third Law ot Motion
More on Vectors
3T
1. Each of the vertically-suspended blocks has the same weight W. The two
forces acting on Block C (Wand rope tension 1) are shown. Draw vectors
to a reasonable scale for rope tensions acting on Biocks A and ~'. ' ••• , " • F
2. The cart is pulled with lorce F at angle e I
as shown. F, and Fy are components of F. I
a. How will the magnrtude of F, change if
the angle IJ is increased by a few degrees?
[more] e[no change]
b. How will the magnrtude of Fy change if
the angle 8 is increased by a few degrees?
e [Iessl (no change]
c. What will be the value of F, it angle 11is OO"?
3T
4. Consider the boom supported by hinge A and by a cable B. Vectors
are shown for the weight Wof the boom at its center. and WI2 for
vertical component of upward force supplied by the hinge
a. Draw a vector representing the cable tension Tat B. Why is it
correct to draw its length so that the vertical component of
T= W/2?
SAME LENGTH AS W/2 GIVES zs; =0,
b. Draw component T,at B. Then draw the horizontal component of
the force at A. How do these horizontal comoonents comoare,
and why EQUAL AND OPPOSITE SO ~Fx :;:O.
...... _- ..... -..... /'l
5. The block rests on the inclined plane. The vector for its weight
W is shown. How many Dther forces act on the block, including
static friction? _~ . Draw them to a reasonable scale.
a. How does the component of W parallel to the plane compare
wrth the force of friction? _~ME SO 1:F< ::.0..: • __ . w
b. How does the component of W perpendicular to the plane compare with the normal force?
SAMESO IFy =O. ..J..••;Jt
-$;,,,,,!
2H
)
CONCEPTUAL PRACTICE PAGE
AppendiX D More About Vectors
Vecrors and sal/boats
(Please do not attempt this until you have studied AppendiX 01)
1. The sketch shows a top view of a small railroad car pulled by a rope. The force Fthat the rope
exens on the car has one component along the track. and another component perpendicular to
the track.
a. Draw these components on the sketch.
Which component Is larger?
PERPENDICULAR COMPONENT
b. Which component produces acceleration?
COMPONENT PARALLEL TO TRACK
c. What would be the effect of pulling on the
rope if rt were perpendicular to the track?
NO ACCELERATION
2. The sketches below represent simplified top views of sailboats in a cress-wind direction.
The impact of Ihe wind produces a FORCE vector on each as shown.
(We do NOT consider velocity vectors here!)
F
a. Why is the position of the sail above
useless for propelling the boat along
its forward direction? (Relate this to
Question l.c above where the train
is constrained by tracks to move in
one direction, and the boat is similarly
constrained to move along one direction
by ne deep vertical fin-the keel.)
AS IN 1.0 ABOVE, THERE'S NO
COMPONENT PABALLEL ra
DIRECTION OF MOTION.
)
b. Sketch the component of force parallel to the
to the direction of the boat's motion (along Its
keel), and the component perpendicular
to its motion. Will the boat move in a forward
direction? (Relate this to Question l.b above.)
2Q
YES, AS IN l,b ABOVE, THERE IS
A COMPONENT PARALLEL TO
DIRECTION TO MOTION.
-il...,ilt .•.t
--J:it~l!.
)