Download newton`s laws

NEWTON'S LAWS
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INTRODUCTION
In the previous chapter we studied the vocabulary and equations that describe motion. Now we will
learn why things move the way they do; this is the subject of dynamics.
An interaction between two bodies, a push or a pull-is called a force. If you lift a book, you exert
an upward force (created by your muscles) on it . If you pull on a rope that's attached to a crate, you
create a tension in the rope that pulls the crate. When a skydiver is falling through the air, the earth
is exerting a downward pull called gravitational force, and the air exerts an upward force called air
resistance. When you stand on the floor, the floor provides an upward, supporting force called the
normal force. If you slide a book across a table, the table exerts a frictional force against the book, so the
book slows down and then stops . Static cling provides a directly observable example of the electrostatic force. Protons and neutrons are held together in the nuclei of atoms by the strong nuclear force,
and radioactive nuclei decay through the action of the weak nuclear force .
The Englishman Sir Isaac Newton published a book in 1687 called The Mathematical Principles of
Natural Philosophy-referred to nowadays as simply The Principia-which began the modern study of
physics as a scientific discipline. Three of the laws that Newton stated in The Principia form the basis
for dynamics and are known simply as Newton's Laws ofMotion .
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THE FIRST LAW
Newton's First Law says that anobject will continue in itsstate ofmotion unless compelled to change bya
force impressed upon it. That is, unless an unbalanced force ads on an object, the object's velocity will
not change: If the object is at rest, then it will stay at rest; and if it is moving, then it will continue to
move at a constant speed in a straight line.
.
Basically, no force means no change in velocity. This property of objects, their natural resistance
to changes in their state of motion, is called inertia. In fact, the First Law is often referred to as the
Law of Inertia.
THE SECOND LAW
Newton's Second Law predicts what will happen when a force does act on an object: The object's
velocity will change; the object will accelerate. More precisely, it says that its acceleration, a, will be
directly proportional to the strength of the total-c-or net-force (Fnet) and inversely proportional to the
object's mass, m:
Fnet =ma
This is the most important equation in mechanics!
The mass of an object is the qua ntitative measure of its inertia; intuitively, it measures how much .
matter is contained in an object. Two identical boxes, one empty and one full, have different masses.
The box that's full has the greater mass, because it contains more stuff; more stuff, more mass. Mass
is measured in kilograms, abbreviated kg. (Note: An object whose mass is 1 kg weighs about 2.2
pounds.) It takes twice as much force to produce the same change in velocity of a 2 kg object than of
a 1 kg object. Mass is a measure of an object's inertia, its resistance to acceleration.
Forces are represented by vectors; they have magnitude and direction. If several different forces
act on an object simultaneously, then the net force, Fnet, is the.vector sum of all these forces. (The
phrase resultant force is also used to mean netforce.)
Since F =ma, and m is a positive scalar, the direction of a always matches the direction of F ~.
~ F = ma, the units for F equal the units of m times the units of a:
Finally, since
-
[F] = [m][a]
= kgm /s'
A force of 1 kg-m/ s' is renamed 1 newton (abbreviated N). A medium-sized apple weighs about 1 N.
THE THIRD LAW
This is the law that's commonly remembered as, to every action, there is an equal, butopposite, reaction.
More precisely, if Object 1 exerts a force on Object 2, then Object 2 exerts a force back on Object 1,
equal in strength but opposite in direction. These two forces, Fl,,,,, and F'.,n.l; are called an action!
reaction pair.
36 •
CRAC KING THE AP: PH YS ICS EXAM
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Examp le 3.1 What net forceis requir ed to maintain a 5000 kg object
moving at a constant velocity of magnitude 7500 ml s?
Solution. The First Law says that any object will continue in its state of motion unless a force acts on
it. Therefore, no net force is requi red to maintain a 5000 kg object moving at a constant velocity of
magnitude 7500 m/ s. Here's another way to look at it: Constant velocity means a = 0, so the equation
F"" =rna immediately gives F"" =0.
Example 3.2 How much force is required 10 cause an object of mass 2 kg
to have an acceleration of 4 ml S2?
Solution. According to the Second Law, F"" = rna = (2 kg)(4 m/s' ) = 8 N.
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:.:, ' :'-': .;~' , .'~'.: Example 3.3 An objectfeels two forces; one of strength 8 ]\1 pulling to the
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left and one of strength 20 N pulling to the right. If the object's mass is 4
; . kg, wha t is its acceleratio n? ' .
;': ' :; ".
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Solution. Forces are represented by vectors and can be added and subtracted. Therefore, an 8 N force
to the left adde d to a 20 N force to the right yields a net force of 20 - 8 = 12 N to the right. Then
Newton's Second Law gives a = F,,/ m = (1 2 N to the right)/(4 kg) =; 3 mi s' to the right.
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WEIGHT
Mass and weight are not the same thing-there is a clear distinction between them in physics-but
they are often used interchangeably in everyday life. The weight of an object is the gravitational force
exert ed on it by the earth (or by whatever planet it happens to be on). Mas s, by contrast, is an intrinsic
property of an object that measures its inertia. An object's mass does not change with location. Put a
baseball in a rocket and send it to the Moon. The baseball's weight on the Moon is less than its weight
here on Earth (because the Moon' s gravitational pull is weaker than the earth' s due to its much
smaller mass), but the baseball's mass would be the same .
Since weight is a force, we can use F = mato compute it. What acceleration would the gravitational
force impose on an object? The gravi tational acceleration, of course! Therefore, setting a = g, the
.
equation F =rna becomes
Fw=mg
This is the equation for the weight of an object of mass m. (Weight is often symbo lized merely by
w, rather than Fw') Notice that mass and weight are propo rtional but not id entical. Furthermore, mass
is measured in kilograms, while weight is measure d in new tons.
Examp le 3.4 What is the mass of an object that weighs 5 00 N?
Solution, Since weight is m multiplied by g, mass is Fw (weight) divid ed by g. Therefore,
m =F j g = (500 N)/(9.8 m/ s') = 51 kg
NEWT ON'S LAW S .
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Example 3.5 A person weighs 150 p ounds. Given that a pound is a un it
of weight equal to.4.45 N, what is this person's mass?
Solu tion. This person's weight in newtons is (150 Ib)(4.45 N /lb) = 667.5 N, so his mass is
m = Fj g = (667.5 N )/ (9.8 m/ s' ) = 68 kg
. ....J':'~. .
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-is 2 kg rests on a. table. Find the
.: '. ':' :,' ,: .:.' magnitude of the force exerted by the table on the book.
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Example 3..6 A book
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Solution. The book experiences two forces: The down ward pull of the earth's gravi ty and the
upward, supporting force exerted by the table. Since the book is at rest on the table, its acceleration
is zero, so the net force on the book must be zero. Therefore the magnitude of the support force mu st
equal the magnitude of the book's weight, which is Fw = mg = (2 kg)(9.8 m/ s') = 20 N.
:.
t,
-..'~:~ ':.-:.- Exampl~ 3.7 A can of paint with ~;~ss of 6 kg hangs from a rope. If the
...
can is to be pulled up to a rooftop with an acceleration of 1 mis', what
,. must the tension in the rope be?
' : - ' . ....
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Solution. First draw a picture. Represent the object of inter est (the can of paint) as a heavy dot, and
draw the forces that act on the object as arro ws connected to the dot. This is called a free-b ody (or
force) dia gram.
(±)
"tip " is the
p ositive direction
11U
T
We draw lots of these.
m
Fw
We have the tension force in the rope, FT (also symbolized merely by T), which is upward, and the
weight, Fw' which is downward. Calling up the positive direction, the n et force is FT - Fw' The Second
Law, F"" = rna, becomes F; - Fw =rna, so
FT = Fw + rna = mg + rna = m(g +a) = 6(9.8 + 1) = 65 N
.::.'..:::;:::,',. .. Exa~pie 3.8 A can of paint with ~ma~~ of 6 kg hangs from a rope. If the
..... .......:'... , can is to be pulled up to arooftop with a constant velocity of1 mis, what
mu st the tension in the rop e be?
.
..<:.;;:::, '
38 •
CRACKING THE AP: PHYSICS EXAM
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Solution . The phrase "constant velocity" automatically means a = 0 and, therefore, F,,, = O. In the
diagram above, FT would need to have the same magnitude as Fw in order to keep the can moving at
a constant velocity. Thus, in this case, FT = Fw = mg = (6)(9.8) = 59 N.
Solution. First draw a free-body diagram
(±)
"up" is the
positive direction
D
U
m
Fw
We have the tension force, FT, which is upward, arid the weight, Fw' which is downward. Calling
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up the positive direction, the net force is FT - Fw . The Second Law, Fnet = ma, becomes FT - Fw =ma, so
FT =Fw + maoRememberi ng that m =Fjg, we find that
F a= 49 N+ 49 N , (9.8m/s' ) =98N
FT =Fw+ ma =Fw +-"'g
9.8m/s
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THE NORMAL FORCE
When an object is in contact with a surface, the surface exerts a contact force on the object. The
component of the contact force that's perpendicular to the surface is called the normal force on the
object. (In physics, the word normal means perpendicular.) The normal force is what prevents objects
from falling through tabletops or you from falling through the floor. The normal force is denot ed by
FN , or simply by N. (If you use the latter notation, be careful not to confuse it with N, the abbreviation
for the newton.)
Example 3.10 A book whose mass is 2 kg rests an ~ table. Find the
magnitude of the normal force exerted by the table on the book.
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Solution. The book experiences two forces: The downward pull of Ear th's gravity and the upward,
supporting force exerted by the table. Since the book is at rest on the table, its acceleration is zero, so
the net force on the book must be zero. Therefore, the magnitude of the support force must equal the
magnitude of the book's weight, which is Fw =mg = (2)(9.8) =20 N. This means the normal force must
be 20 N as well: FN = 20 N. (Notice that this is a repeat of Example 3.6, except now we have a name
. for the "upward, supporting force exerted by the table"; it's called the normal force.)
NEWTON 'S LAWS .
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FRICTION
When an object is in contact with a surface, the surface exerts a contact force on the object. The
component of the contact force that' s parallel to the surface is called the friction force on the object.
Friction, like the normal force, arises from electrical interactions between atoms that comprise the
object and those that comp rise the surface.
We;lliook at two main categories of friction: (l) static friction and (2) kinetic (sliding) friction.
If you attempt to pus h a heavy crate across a floor, at first you meet with resistance, but then you pu sh
hard enough to get the crate moving. The force that acted on the crate to cancel out your initial pushes
was static friction, and the force that acts on the crate as it slides across the floor is kinetic friction.
Static friction occurs when there is no relative motion between the object and the surface (no sliding);
kinetic friction occurs whe n there is relative motion (when there's sliding).
The strength of the friction force depends, in general, on two things: The nature of the surfaces
and the strength of the normal force. The nature of the surfaces is represented by the coefficient of
friction, denoted by 11 (mu ). The greater this number is, the stronger the friction force will be. For
example, the coefficient of friction between ru bber-soled shoes and a wooden floor is 0.7,but between
rubbed-soled shoes and ice, it's only 0.1. Also, since kinetic friction is generally weaker than static
friction (it's easier to keep an object slidi ng once it's sliding than it is to start the object sliding in the
first place), there are two coefficients
of friction; one for static friction (11 ,) and one for kinetic friction
.
(Ilk)' For a given pair of surfaces, it's virtually always true that Ilk < 11 s' The strengths of these two
types of friction forces are given by the following equations:
Fstatic friction, max =
Jil
N
Fkinetic friction = )1kFN
Notice that the equation for the strength of the static friction force is for the maximum value only.
This is because static friction can vary, precisely counteracting weaker forces that attemp t to move an
object. For example, suppose an object feels a normal force of FN = 100 Nand the coefficient of static
friction between it and the surface it's on is 0.5. Then, the maximum force that static friction can exert
is (0.5)(100 N) =50 N. However, if you push on the object with a force of, say, 20 N, then the static
friction force will be 20 N (in the opposite direction), not 50 N; the object won' t move. The net force
on a stationary object must be zero. Static friction can take on all values, up to a certain maximum,
and you must overcome the maximum static friction force to get the object to slide. The direction of
Fkinotidri<tion =Ff(~nclid is opposite to that of moti on (sliding), and the direction of F" atic friction =Ff(""") is
opposite:to that of the intend ed motion.
Example 3.11 A crate of mass 20 kg is sliding across 'a wooden floor. The
coefficient of kinetic friction between the crate and the floor is 0.3. .;-. " •.
'(a) Determine the strength of the friction force acting on the crate. .
(b) If the crate is being pulled by a force of 90 N (parallel to the floor),
find the acceleration of the crate.
Solution. First draw a free-body diagram:
40 •
CRACK ING THE AP : PHYSICS EXAM
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If an object
doesn't move
than the force
exerted is the
static frictional
force.
Free Body Diagram
Fw
(a) The norm al force on the object balances the object's weight, so
FN = mg =·(20 kg)(9.8 m/ s' ) = 196 N.therefore, F'lki"'"cl = J.l,FN = (0.3)(196 N) = 59
N.
F, = 90 N - 59 N = ~1 N, so the
acceleration of the crate is a = Fn"lm = (31 N)/(20 kg) = 1.6 m/ s"
(b) The net horizontal force that acts on the crate is F -
·:'r~\'?~)\\:' Eic~ple 3;12- A ~;~ of ~~100 kg'r~~ on th; fl09;. Th~·~~ffici;n~~f·~;·. : l:;;::;\:;:,
..
::,:, :~: ~. :y,~., ~ static friction is 0.4. If a force of 250 N (parallel to the floor) is applied to , .... '. :.';:.:
' ':i.:~'.: ,': .::"
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the crate, what's the magnitude of the force of static friction on the crate? ..
.:...:.~~
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Solution. The normal force on the object balances its weight, so FN = mg = (100 kg)(9.8 m/ s' ) = 980 N.
""IC). max = r'/I sFN = (0.4)(980 N) = 390 N . This is the maximum
force that static
Therefore, Fs,ati'cfrict'
ion-max = F" s...
.
friction can exert, but in this case it's not the actual value of the static friction force. Since the appli ed
force on the crate is only 250 N, which is less than the F fl"'tid. """, the force of static friction will be less
also: F' '''''tid = 250 N, and the crate will not slide.
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PULLEYS
Pulleys are devices that change the direction of the tension force in the cords that slide over them .
Here we'll consider each pulley to be frictionless and massless, which means that their masses are so
much sma ller than the objects of interest in the problem tha t they can be ignored .
pulley
f-------~I
m
M
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~
~
";,
~
.-......
._.
~...
"
._ .--:
_.
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;::;;," >';.;::; : Example 3.13.Inthe diagram above, assume thafthe tabletop is frictionless.
.
\:;"::\:;/~~ Determine the acceleration of the blocks once they're released from rest. ~:':"'"
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.:,ok.o,;;r.
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NEWTON'S LAWS .
41
Solutio n. There are two blocks, so we draw two free-body diagrams:
:-+__FT
m
D
M
To get the acceleration of each one, we use Newton's Second Law, F"" =rna.
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,
,
.
c=~> 0 \ h h. hh Oo Oo OohOohhOo .h u+ F = rna
T
,
~.-~~
,,,
,
m
:-+---+ JFT
'-'" ,- . ' . ...M g-FT =Ma
,
,
,
,
:
,
M
,
D
.
,
0/
42 •
CRACKING THE AP: PHYSICS EXAM
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Notice that there are two unknowns, FT and a, but we can eliminate FT by adding the two
equations, and then we can solve for a,
FT=rna}
Mg - FT = Ma
; ) Add theequations
'to eliminate FT,
Mg'=rna+Ma
= a(rn+M)
Mg
rn+M
fry;:;:r:, f7, :;~:-:P- -,"'Bi?"f" ~~-
a
_ ~P~ J"'-"""I1--;::; ';Y ' 'H?
-.
0"_~1''':Y~ ~<1
-~'1 ' "l-e ~4t~ 'r:<_·"''Pn ;:'Wi' ~~ " ~~,~,:'1i!" ~~"~: """
!;~{~;; ~;;,~r;~ Example 3,t4. Using the saniediagFam <l$ In tlte pr~Mi€ll,lS ~am)i1le, as- . :
~~~~ i}~;E~;f:! sume that m = 2. kg, 111 = 10 kg, and the"coeff!icient of kinetic friction "
i~;~r,;;;tS~' \~~;, between the small blo~kandthe tabl~top is O,li.Compute t~e acceleration :
t
>:
~i~~i~}h1~~;E~~~~~f· ~~~~~~~;~:t~i21~W~k~n~~~-~t~~;·::~U\1:;r~\3~~~-~~:~:;t[liib~~,~~t~;:,:;E,1:,;j~;a
Solution. Once again, draw a free-body diagram for each object, Notice that the only difference
between these diagrams and the ones in the previous example is the inclusion of the force of (kinetic)
friction, Fr that acts on the block on the table.
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m
,
,
,,
,,,
M
,,
,
.
,,
,
0:
,
"
Fw= Mg ,/
As before, we have two equations that contain two unknowns (a and FT) :
FT - F, = rna
(1)
NEWTON'S LAWS .
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Mg -FT=Ma (2)
Add the equations (thereby elimina ting FT) and solve for a. Notice that, by definition, Fr"= /l FN,
and from the free-body diagram for m, we see that FN =mg, so Ff = /lmg:
Mg - Ff = ma + Ma
Mg -/l mg = a(m + M)
M -Ilm
m+M
Substituting in the numerical values given for
g =a·
m, M, and u, we find that a = ~g (or 7.4 m /s').
.:. : Example 3.15 In the previous example, calculate the strength of the
-r. , tension in the cord.
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Solution. Since the value of a has been determined, we can use either of the two original equations
to calculate FT. Using Equation (2), Mg - FT= Ma (because it's simpler), we find
3
1
1
FT = Mg - Ma =Mg -M ·-g= -Mg= -(10)(9.8)= 25 N
4
4
4
As you can see, we would have found -the same answer if Equation (1) had been used:
FT
J
= Ff + ma= j.nng + rna = j.nng + m. . ~4 g = rng(I.t+~)
4
= (2)(9.8)(0.5 +0.75)
= 25 N
INCLINED PLANES
An in clined plane is basically a ram p. If an object of mass mis on the ramp, then the force of gravity
on the object, Fw· = mg, has two componen ts: One that's parallel to the ram p (mg sin e) and one that's
normal to the ramp (mg cos e), where e is the incline angle. The force driving the block down the
inclined pla ne is the component of the block's weigh t that's parallel to the ramp: mg sin e.
KNOW THIS EQUATION.
KNOW THIS EQUATION.
44 •
CRACKING THE AP: PHYSICS EXAM
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.: ).: Example 3.16' A block slides down a frictionless, inclined plane that
" '.: makesa 30° angle with
the horizontal.
Findthe accelerationof thisblock. ..........ii:.:-""'"
....... -o.;' _",...
.....
"'""'_ -"'
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Solution. Let m denote the mass of the block, so the force that pulls the block down the incline is
e, and the block's acceleration down the plane is
mg sin
mg sin e ,
1
gsin e = gsin30' = - g = 4.9 m / s'
m
m
2
F
a= -
Example 3.17 A block slides down an inclinedplane that makes a 30°
.:: angle with the horizontal. If the coefficient of kinetic friction is 0.3, find
:';.:·:;~·{~1'·" the aeceleration of the block. .' ,',', ",:t. ,: ', . ' "
.:t. ..', : : ".
Solution. First draw a free-body diagram. Notice that, in the diagram shown below, the weightof the
block, Fw = mg, has been written in terms of its scalarcomponents:Fw sin 9 parallel to the ramp and
Fw cos e normal to the ramp:
The force of friction,Ff , that acts up the ramp (opposite to the direction in which the blockslides)
has magnitudeFf = jlF w Butthe diagram shows thatF N =Fw cos e, so F, = jl (mg cos 9). Therefore the
net force down the ramp is
Fw sin e - Ff = mg sin e - jlmg cos e = mg(sin e - jl cos 9 )
Then, setting F"'. equal to rna, we solve for a:
F ..
a= - n
mg(sine- jlcose)
m
m
=' g(sin e-
jl cose)
= (9.8 m / s')(sin30' - 0.3cos30')
= 2.4m /s'
NEWTON'S LAWS .
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UNI FORM CIRc:U LAR MOTION
)
In Chapte r 2, we considered two types of motion; straight-line motion and parabolic motion. We will now
look at motion that follows a circular path, such as a rock on the end of a string, a horse on a merry,goround, and (to a good approximation) the Moon around Earth and Earth around the SUn.
Let's simplify matters and consider the object's speed around its path to be constant. This is called
uniform circular motion. You should remember that although the speed may be constant, the velocity
is no t, because the direction of the velocity is always changing. Since the velocity is changing , there
must be acceleration. This acceleration do es not change the speed of the object; it only changes the
direction 'of the velocity to keep the object on its circular path. Also, in order to produce an acceleration,
there must be a force; otherwise, the object would move off in a straight line (Newton' s First Law).
The figu re on the left below shows an object moving along a circular trajector y, along with its
velocity vectors at two nearby points. The vecto r VI is the object's velocity at time I = I I' and v, is the
object's velocity vector a short time lat~r (at time I = I') . The velocity vector is alw ays tangential to the
object's path (whatever the shape of the trajectory). No tice that since we are assuming cons.tant speed,
the leng ths of V I and v, (their magnitudes) are the same .
Since Ii v =v, - VI points tow ard the center of the circle (see the figur e on the right ), so does the
acceleration, since a = Ii v I M . Because the acceleration vector points toward the center of the circle,
it' s called centripetal acceleration, or a" The centripetal acceleration is wh at turns the velocity vector
to keep the object traveling in a circle. The magnitude of the cent ripetal acceleration depends on the
object's speed, v, and the radiu s, r, of the circular path according to the equation
v'
a=, r
. "0'
,
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Example 3.18 An object of mass 5 kg moves at a constant speed of
., 6 ml s in a circular path of radius 2 m . Find the object's acceleration and
the net force responsible for its mo tion. '
.".
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. ,.
,
.
Solution. By definition, an object moving at constant speed in a circular p ath is undergoin g uniform
circular motion . Therefore, it experiences a centripetal acceleration of m agnitude ifl r, always directed toward th e center of the circle:
v2
r
a =c
(6 m /s)'
2m
.
18 mis'
The force that produces the centripetal acceleration is given by New ton's Second Law, coupled
with the equation for centripe tal acceleration:
v'
FC =maC =m-r
46 • CRACKIN G THEAP : PHYSICS EXA M
F = ma
IMPORTANT!
This equation gives the magnitude of the force. As for the direction, recall that because F =rna, the
directions of F and a are always the same. Since centripetal acceleration points toward the center of
the circular path, so does the force .that produces it. Therefore; it's called centripetal force. The
centripetal force acting on this object has a magnitude of F, =rna, =(5 kg)(18 m/s') =90 N.
)
Example 3.19 An 8.3 kg mass is attached.to a string that has a breaking
strength of HiOO N. If the mass is whirled in a horizontal circle of radius
.' '.
',, .
80 em, what maximum speed can it have? . '
, \"
Solution. The first thing to do in problems like this is to identify what forcets) provide the centripetal
force. In this example, the tension in the string provides the centripetal force:
(0.80 m)(1500 N)
8.3 kg
= 12 rn/ s
)
,
.
.',
"
Example 3.20 An athlete who weighs 790N is running around a curve at
a speed of 6.0m/s in an arc whose radius of curvature, r, is 4.9 m. Find
the centripetal force acting on him. What provides the centripetal force? .
What could happen to him ifr were smaller?
Solution. Using the equation for the strength of the centripetal force, we find that
F
c
2
= m v = Fw
r
g
.
v
2
r
=
790 N , (6.0rn/s)2
9.8N/kg
4.9 rn
590N
In this case, static friction provides the centripetal, force. Since the coefficient of static friction
between his shoes and the ground is most likely around 1, the maximum force that static friction can
exert is /llN ~ FN= Fw = 790 N. Fortunately, 790N is greater than 590 N.But notice that ifthe radius
of curvature of the arc were much smaller, then F, would become greater than what static friction
could handle, and he would slip.
"
, .';
Example 3.21 A roller-coaster car enters the circular-loop portion of the
ride . At the very top of the circle (where the people in the car are upside
down), the speed of the car is 25 m/ s, and the acceleration points straight
down. If the diameter of the loop is 50 m and the tota l mass of the car
(plus passengers) is 1200 kg, find the magnitude. of the normal force
exerted by the track on the car at this point.
«
-NEWTON'S LAWS .
47
Solution. There are two forces acting on the car at its topmost point: The normal force exerted by the
track and the gravitational force, both of which point downward .
The combination of these two forces, FN + Fw' provides the centripetal force:
mv
=--mg
r
=m(vr _g)
2
2
=(1200
kg)[ (~'2(50m /sm))2 .
2
9.8 m /S ]
=1.8x10' N
~,
·r·
. ,. . EXample 3.22 In the previous example, if the net force on the car at its
.. " . topmost paint is straight down, why doesn't the car fall straight down?
----
Solu tion. Remember that force tells an object how to accelerate. If the car had zero velocity at this
point, then it would certainly fall straight down, but the car has a nonzero velocity (to the left) at this
poin t. The fact that the acceleration is downward means that, at the next moment v will point down
to the left at a slight angle, ensuring that the car remains on a circular path, in contact with the track.
48 • CRAC KING THEAP : PHYSICS EXAM
)
CHAPTER 3 REVIEW QUESTIONS
)
SEGIONI: MU lTI PLE CHOICE
l. A person standing on a horizontal floor
feels two forces: the downward pull of
.gravity and the upward supporting force
from the floor . These two forces
5. The coefficient of static friction between a
box and a ramp is 0.5. The ramp's incline
angle is 30' . If the box is placed at.rest on
the ramp, the box will
(A) have equal magnitudes and form an
action/reaction pair.
(B) have equal magnitudes but do not
form an action/reaction pair.
(C) have unequal magnitudes and form
an action/reaction pair . .
(D) have unequal magnitudes and do not
form an action/reaction pair.
(E) None of the above
(A) accelerate down the ramp.
(B) accelerate briefly down the ramp but
then slow down and stop.
(C) move with constant velocity down
the ramp .
(D) not move.
(E) Cannot be determined from the
information given"
6.
2. A person who weighs 800 N steps onto a
scale that is on the floor of an elevator car.
If the elevator accelerates upward at a rate
of 4.9 m/ s', what will the scale read?
,
•
(A) 400N
(B) 800 N
(C) 1000 N
(D) 1200 N
(E) 1600 N
3. A frictionless inclined plane of length
20 m has a maximum vertical height of
5 m. If an object of mass 2 kg is placed on
the plane, which of the following best
approximates the net force it feels?
I
,.
(A) 5 N
(8) lO N
(C) 15 N
(D) 20 N
(E) 30 N
Assuming a frictionless, massless pulley,
determine the acceleration of the blocks
on ce they are released from rest.
(A)
4. A 20 N block is being pushed across a
horizontal table by an 18 N force. If the
coefficient of kinetic friction between the
block.and the table is 0.4, find the accel eration of the block.
(A) 0.5 m/s'
(B) 1 m/s'
5 m/s'
(D) 7.5 m/ s'
(E) 9 m/s'
«»
(B)
(C)
m
g
M+m
M
M +m
g
M
-g
m
M +m
(D)
g
M -m
M-m
(E)
g
M+m
NEWTON'S lAWS .
49
7. If all of the forces acting on an object
balance so that the net force is zero, then
(A) the object must be at rest.
(B) the object's speed will decrease.
(C) the object will follow a parabolic
trajectory.
(0) the object's direction of motion can
change, but not its speed.
(E) None of the above
8. A block of mass m is at rest on a frictionless, horizontal table placed in a laboratory on the surface of the earth. An
identical block is at rest on a frictionless,
horizontal table placed on the surface of
the Moon. Let F be th e net force necessary
to give the Earth-bou n d block an acceleration of a across the table. Given that g Moon
is one-sixth of gEarth the force necessary to
give the Moon-bound block the same
acceleration a across the table is
.
l
(A) F / 12
(B) F/6
(C) F/ 3
(0) F
(E) 6F
9. A crate of mass 100 kg is at rest on a
horizontal floor. The coefficient of static
friction between the crate and the floor is
0.4, and the coefficient of kinetic friction is
0.3. A force F of magnitude 344 N is then
applied to the crate, parallel to the floor.
Which of the following is true?
.
(A) The crate will accelerate across the
floor at 0.5 m/s'.
(B) The static friction force, which is the
reaction force to F as guaranteed
by Newton's Third Law, will also
have a magnitude of 344 N .
(C) The crate will slide across the floor at
a constant speed of 0.5 m/ s.
(0) The cra te will not move.
(E) None of the above
50 • CRA CKING THE AP : PHYSICS EXAM
10. Two crates are stacked on top of each
other on a horizontal floor; Crate #1 is on
the bottom, and Crate #2 is on the top.
Both crates have the same mass. Com pared to the strength of the force F,
necessary to push only Crate #1 at a
constant speed across th e floor, the
strength of the force F, necessary to push
the stack at the same constant speed
across the floor is greater than F1 because
(A) the force of the floor on Crate #1 is
greater.
(B) the coefficient of kinetic friction
between Crate #1 and the flooris
greater. .
(C) the force of kinetic friction, bu t not
. the normal force, on Crate #1 is
.
greater.
(0) the coefficient of static friction
between Crate #1 and the floor is
greater.
(E) th e weight of Cr ate #1 is greater.
11. An object moves at constant speed in a
circular path. Which of the followmg
statements is/ are true?
I. The ve locity is constant.
II. Th e acce leration is constant.
III. The net force on the object is
zero since its speed is constant.
(A) II only
(B) I and III only
(C) II and III only
(0) I and II only
(E) None of the above
Questions 12 13:
A 60 em rope is tied to the handle of a bucket
which is then whirled in a vertical circle . The
mass of the bucket is 3 kg .
12. At the lowest poin t in its path, the tension
in th e rope is 50 N . What is th e speed of
the bucket?
14. An object moves at a constant speed in a
cir cular path of radius r at a rate of 1
revolution per second. What is its accel.eration?
(A)
(B)
(C)
(D)
(E)
a
(27<'
(27< '
(47<'
(47<'
s-')r
s-')r'
s-')r
s-')r'
mls
mls
mls
(D) 4 mls
(E) 5 mls
(A) 1
(B) 2
(C) 3
13. What is the critical sp eed below which the
rope would become slack when the bucket
reaches the highest pointin the circle?
(A) 0.6 mls
(B) 1.8 mls
(C) 2 .4 mls
(D) 3.2 mls
(E) 4.8 mls
)
NEWTON 'S LAWS .
51
SEGION II: FREE RESPONSE
)
a
1. This question concerns the motion of crate being pulled across a horizontal floor by a rope.
In the diagram below, the mass of the crate is m, the coefficient of kinetic friction between the
crate and the'flooris u , and the tension in the rope is FT'
(a)
Draw and label all of the forces acting on the crate.
Cb)
Compute the normal force ac ting on the crate in terms of m, FT , 8, and g.
Cc)
Compute the acceleration of the crate in terms of m, FT , 8, u, and g.
(d)
[Cj Assume that the magnitude of the tension in the rope is fixed but that the angle 8
may be varied. For what value of 8 would the resulting horizontal acceleration of the
" crate be maximized?
2. In the diagram below, a massless string connects two blocks-of masses m1 and m21 respectively-on a fla t, frictionless tab letop. A force F pulls on Block #2, as shown:
Block #1
Block #2
___...J[:J---15]- - . .
• F
(a)
Draw and label all of the forces acting on Block #1.
(b)
Draw and label all of the forces acting on Block #2.
(c)
What is the acceleration of Block #1?
(d)
What is the tension in the string connecting the two blocks?
(e)
If the string connecting the blocks were 'n ot massless, but instead had a mass of m,
figure out
(i)
(ii)
52 •
the acceleration of Block #1, and
the difference between the strength ofthe force that the connecting string exerts on
Block #2 and the strength of the force that the connecting string exerts on Block #1.
CRACKIN G THE AP: PHYSIC S EXAM
3. In the figure shown, assume that the pulley is frictionless and massless.
)
(a)
If the surface of the inclined plane is frictionless, determine what valuers) of "8 will
cause the box ofmass mt to
(i) accelerate up the ramp;
(ii) slide up the ramp at constant speed.
(b)
If the coefficient of kinetic friction between the surface of the inclined plane and the box
of mass m, is Ilk' derive (but do not solver .an equation satisfied by the value of e
which will cause the box of mass m, to slide up the ramp at constant speed.
4. A ~ky diver is falling with speed V o through the air . At that moment (time t = 0), she opens her
parachute and experiences the force of air resistance whose strength is given by the equation
F = ku, where k is a proportionality constant and v is her descent speed. The total mass of the
sky diver and equipment is m. Assume that g is constant throughout her descent.
)
,
(a)
Draw and label all the forces acting on the sky diver after her parachute opens.
(b)
Determine the sky diver's acceleration in terms of m, v, k, and g.
(c)
Determine the sky diver's terminal speed (that is, the eventual constant speed of
descent).
(d)
Sketch a graph of v as a function of time, being sure to label important values on the
vertical axis.
(e)
[C] Derive an expression for her descent speed, v, as a function of time t since opening
her parachute in terms -of m, k, and g.
I
NEWTON'S LAWS .
S3
5, An amusement park ride consists of a large cylinder that rotates around its central axis as the
passengers stand against the in n er wall of the cylinder. Once the passengers are moving at a
certain speed v, the floor on which they were standing is lowered. Each passenger feel s
pinned against the wall of the cylinder. as it rotates . Let r be the inner radius of the cylinder.
(a)
Draw and label all the forces acting on a passenger of mass rn as the cylinder rotates
with the floor-lowered.
(b)
Describe wha t conditions must hold to keep the passengers from sliding down the wall
of the cylinder. '
(c)
Compare the conditions discussed in part (b) for an adult passenger of mass m and a
child passenger of mass m/2.
.
6. A curved section of a highway has a radius of curvature of r. The coefficient of friction
between standard automobile tires and the surface of the highway is 11,
(a)
Draw and label all the forces acting on a car of mass m traveling along this curved part
of the highway.
(b)
Compute the maximum speed with which a car of mass m could make it around the
turn without skidding in terms of J1." r, g, and m.
City engineers are planning on banking this curved secti on of highway.at an angle of
ho rizontal.
(c)
Draw and label all of the forces acting on a car of mass m traveling along th is banked
turn. Do not include friction.
(d)
The engineers want to be sure that a car of mass m traveling at a constant speed v (the
posted speed limit) could m ake it safely around the banked turn even if the road were'
covered with ice (that is, essentially frictio nless). Compute this banking angle in
terms of r. v, g, and m.
e
54 •
e.to the
CRACKING THE AP : PHYS ICS EXAM
)
Thus, the cannonball's vertical position can be written in terms of its horizontal position as
follows:
(2)
Substituting the known values for x, ()0' g, and
DO'
y(at x = 220 m) = (220 m)tan 40°
we get
(9.8 m/ s')(220 m)2
2(50m / s? cos' 40°
= 23 m
This is indeed less than 30 m, as desired.
(b) From Equation (1) derived in part (a),
220 m
=5.7 s
(50 m / s) cos 40°
(c) The height at which the cann onball strikes the wall was determined in part (a) to be 23 m.
)
(d) We use Equation (2) from part (a):
y = x tan ()o
Using the identities 1/cos' ()0 = sec' ()0 and sec' ()0 = 1 + tan' ()0' this equat ion can be written in
the form
or
Substituting x = 220 m and y = 30 m (which is the top of the castle wall), this equation becomes
(94.864)(ta n()o)' - 220 tan ()o + 124.864 = 0
SOLUTIONS TO THE CHAPTER REV IEW QUESTIONS .
431
)
The quadratic formula then gives
tane =
o
eo
220± ~2202 -4(94.864)(124.864)
2(94.864)
= 0.9913, 1.328
= 44.7°, 53.0°
4. (a) Integrating a(t) with respect to time gives the velocity, v(t):
v(t) = f art)dt = f 6t dt =3t 2 + V o
2
=}
v(t) =3t + 2
Setting this equal to 14, we solve for t:
v(t)=14
=}
3t 2 +2= 14
=}
t=2s
(Notice that we discarded the solution t = -2.)
(b) Integrating v(t) with respect to time gives the position, x(t):
x(t) = fv(t)dt = f(3t 2+2)dt=t 3+2t +xo
=}
x(t)=e+2t+4
Therefore, the particle's position at t =3 s is
x(3) = [t 3 + 2t +41..3 = 37 m
CHAPTER 3 REVIEW QUESTIONS
SEGION I: MUlTIPLE CHOI CE
1. (B) Because the person is not accelerating, the net force he feels must be zero. Therefore, the
magnitude of the upward normal force from the floor must balance that of the downward
gravitational force. Although these two forces have equal magnitudes, they do not form an
action/reaction pair because they both act on the same object (namely, the person) . The forces
in an action/reaction pair always act on different objects.
432 • CRAC KING THE AP PHYSICS EXAM
}
2. (D) First dr aw a free-body diagram:
(f)
thJI
"u p " is
p ositive d irection U
m
Fw
The person exerts a downward force on the scale, and the scale pushes up on the person with an
equal (but opposite) force, FN ' Thus, the scale reading is FN' the ma gnitude of the normal force.
Since FN - Fw =rna, we have FN =Fw + rna =(800 N) + [800 N/(9.8 m/ s')](4.9 m/s') =1200 N.
3. (Al The net force that the object feels on the inclined plane is mg sin (J, the component of the
gravitational force that is parallel to the ramp . Since sin (J= (5 m )/ (20 m) = 114, we have
F"" = (2 kg)(9.8 N / kg)(1/ 4) =5 N.
4. (C) The net force on the block is F - F, =F- fl. ! N =F - fl. ! w =(18 N) - (0.4)(20 N) =10 N. Since
FM
' =rna =(Fjg)a, we find that 10 N =[(20 N)/ (9.8 m/s' )]a, which gives a =5 m / s'.
)
rng sin (J, and the maximum force of static
friction is fl. l N = fl. ,rng cos (J . If rng sin (J is greater than fl. ,rng cos (J, then there is a netforce
down the ramp, and the block will accelerate down . So, the questi on becomes, "Is sin (J greater
than fl. , cos (J?" Since (J = 30° and fl. , = 0.5, the answer is yes.
5. (A) The force pulling the block down the ramp is
6. (E) One wa y to attack this question is to notice that if the two masses happen to be equal, that
is, if M = m, then the blocks won't accelerate (because their weights balance). The only
expression given that becomes zero when M = rn is the one given in choice (E). If we draw a
free-body diagram,
FT
(f)
the11
"up" is
positive direction (for m)
U
m
M
n "d own" is the
JJ p o sitiv e d irection (for M)
(f)
mg
Mg
'SOLUTI ONS TO TH E CHAPTER REVIEW QUESTIONS .
433
Newton's Second Law gives us the following two equations:
FT-mg =ma
Adding these equations yields Mg - mg =ma + Ma
a
Mg-mg
M+m
(1 )
= (M + m)a, so
M-m
g
M+m
7. (E) 1f Fre t = 0, then a = 0. No acceleration means constant speed (possibly, but not necessarily,
zero) with no change in direction. Therefore, statements B, C, and 0 are false, and statement
A is not necessarily true.
8. (0) The horizontal motion across the frictionless tables is unaffected by (vertical) gravitational
acceleration. It would take as much force to accelerate the block across the table on Earth as it
would on the Moon . (If friction were taken into account, then the smaller weight of the block
on the Moon would imply a smaller normal force by the table and hence a smaller frictional
force. Less force would be needed on the Moon in this case.)
9. (0) The maximum force which static friction can exert on the crate is Ji.,FN = Ji. ,Fw = Ji. ,mg =
(0.4)(100 kg)(9.8 N/kg) = 392N. Since the force applied to the crate is only 344 N, static friction
is able to apply that same magnitude of force on the crate, keeping it stationary. [Choice (B) is
incorrect because the static friction force is not the reaction force to F; both F and F, (,t"'1 act on
the same object (the crate) and therefore cannot form an action /reaction pair.]
10. (A) With Crate #2 on top of Crate #1, the force pushing downward on the floor is greater, so the
normal force exerted by the floor on Crate #1 is greater, which increases the friction force.
Statements B, C, 0, and E are all false.
11. (E) Neither the velocity nor the acceleration is constant because the direction of each of these
vectors is alwa ys changing as the object moves along its circular path. And the net force on the
object is not zero, because a centripetal force must be acting to provide the necessary centripetal
acceleration to maintain the object's circular motion.
12. (B) When the bucket is at the lowest point in its vertical circle, it feels a tension force FT upward
and the gravitational force Fw downward. The net force toward the center of the circle, which
is the centripetal force, is FT - Fw' Thus,
F - F = m v'
T
w
=> v = t (FT -mg) = (0.60 m)[50 N -(3 kg)(9.8 N / kg)] 2 m / s
r
m
3kg
13. (C) When the bucket reaches the topmost point in its vertical circle, the forces acting on the
bucket are its weight, Fw' and the downward tension force, FT' The net force, Fw + FT, provides
the centr ipetal force. In order for the rope to avoid becoming slack, FT must not vanish.
Therefor e, the cut-off speed for ensuring that the bucket makes it aroun d the circle is the speed
434 •
CRA CKIN GTHE AP PH YSICS EXAM
)
at which FT just becomes zero; any greater speed would imply that the bucket would make it
around. Thus,
Vcut.off =
fTF: r;:;
v--;;;
= "gr
=
~r(9-.8-m-l---'s2::-)(-O.-60-m-)
=2.4m / s
14. (D) Centripetal acceleration is given by the equati on a, = v2 l r. Since the object covers a distance
of 2 rtr in 1 revolution, its speed is 2 tcr S-I. Therefore,
SEGIONII: FREE RESPONSE
1. (a) The forces acting on the crate are FT (the tension in the rope), Fw (the weight of the block),
FN (the normal force exerted by the floor), and F, (the force of kinetic friction):
FT
)
(b) First break FT into its horizontal and vertical components:
Ff _ -.J--+.:+----'---- - -4'
SOLUTI ONS TO THE CHA PTER REV I EW QUEST IO NS .
435
Since the net vertical force on the crate is zero, we have FN + FT sin 8
FTsin 8 =mg-FTsin 8.
= Fw'
so FN
= Fw -
)
(c) From part (b), we see that the net horizontal force acting on the crate is
FTcos8- Ff = FTcos8-J.LfN = FTcos8-lJ.(mg- FTsin8)
so the crate's horizontal acceleration across the floor is
a= F,,, = F,cos8-Il(mg-F,sin8)
m
m
(d) In order to maximize the crate's acceleration, we want to maximize the net horizontal force,
F,cos8-Il(mg-F,sin8) = FT(cos8+ Ilsin8)-llmg. Since FT, u, m, and g are all fixed,
maximizing the net horizontal force depends on maximizing the expression
cos8+ jlsin8
Call this I( 8). To determine the value of 8 at which 1(8) = cos8 + jl sin 8 attains an extreme
value, we take the derivative of 1and set it equal to zero:
,,'
f' (8) = - sin 8 + jlcos8 = 0
jlcos8 = sinO
jl = tan 8
8=tan-1 jl
[That this value of 8 does indeed maximize 1can be verified by noticing that for 8 in the
interval 0 ~ 8 ~ ~ 1t, r(8) = -cos8- jlsin8 is always negative, and
I( tan
-1
lJ.) = cos( tan - 1 lJ.)+ usin] tan - 1 lJ.)
_
1
+lJ. lJ.
- ~1+lJ.2
~1+lJ.2
=~1+lJ.2
is greater than 1(0)
= 1 or 1(~ 1t) = u, the values of 1at the endpoints of the interval.]
436 • CRACKING THE AP PHYSICS EXAM
}
2. (a)The forces acting on Block #1 are FT (the tension in the string connecting it to Block #2), Fwl
(the weight of the block), and FNI (the normal force exerted by the tabletop):
FNI
H----~FT
FW1
(b) Theforces acting on Block #2are F (thepulling force), FT (the tensionin the stringconnecting
it to Block #1), Fw2 (the weight of the block), and FN, (thenormal force exertedby the tabletop):
\
I
FT+ - - ---I--!-I-- - - - F
Fw 2
(c) Newton's Second Law applied to Block #2yields F-FT = m,a, and applied to Block #1 yields
FT = mp. Adding these equations, we find that F = (m l + m,)a, so
(d) Substituting the result of part (c) into the equation FT =mIa , we get
SOLUTIONS TO THE CHAPTER REVIEW QUESTIONS .
437
(e) (i) Since the force F must accelerate all three masses-mj' m, and m,-the common acceleration of all parts of the system is
F
a
(e) (ii) Let Fn denote the tension force in the connecting string acting on Block #1, and let F1'2
denote the tension force in the connecting string acting on Block #2. Then, Newton's Second
Lawapplied to Block #1 yields Fri = mja and applied to Block #2 yields F- F1'2 = m,a. Therefore,
using the value for a computed above, we get
Frz - Fn
=(F -m2a) -m,a
=F-(m, +m 2)a
=F-(m, +m 2 ) -
-
F
-
-
m, +m+m2
=F(l-
m~::~J
m
=F - - - m, +m+m2
3. (a) First draw free-body diagrams for the two boxes:
Applying Newton's Second Law to the boxes yields the following two equations:
F; - m,g sin
438 • CR ACKING THE AP PHYSI CS EXAM
= mja
(1)
m,g - F; = m,a
(2)
(J
Adding the equations allows us to solve for a:
m,g - m,gsinO= (m, + m2 )a
m -m,sinO
a= 2
g
m, +m2
(i) For a to be positive, we must have m, - m, sin 0 > 0, which implies that sin 0 < m,lm" or,
equivalently, 0 < sin-'(m,l m,).
(ii) For a to be zero, we mus t have m, - m, sin 0 = 0, which implies that sin 0
equivalently, 0 =sin-'(m,lm,).
=m,lm" or,
(b) Including the force of kinetic friction, the force diagram for m, is
)
Since F, = III
equations:
N
= Ilkm ,g cos 0, applying Newton's Second Law to the boxes yields these two
Adding the equations allows us to solve for a:
m,g-m,gsin O- f.l..m,gcosf! = (m, +m., )a
a= m, - m,(sin f! +f.l.. cosO) g
m1+m2
If we want a to be equal to zero (so that the box of mass m, slides up the ramp with constant
velocity), then
m, - m,(sin O+ Ilkces s) =
°
sin f!+f.l.. cos f! = m,
m,
SOLUTIONS TO THE CHAPTER REVIEW QUESTIONS .
439
4. (a) The forces acting on the sky diver are F~ the force of air resistance (upward), and Fw' the
weight of the sky diver (downward):
F,
sky diver
(mass = m)
Fw
(b) Since Fn "
=Fw -
F, =mg- ko, the sky diver's acceleration is
a= Fnet
m
= mg-kv
m
(c) Terminal speed occurs when the sky diver's acceleration becomes zero, since then the
descent velocity becomes constant. Setting the expression derived in part (b) equal to 0, we
find the speed v = v, at which this occurs:
v=v, whena=O
=;
mg-kv,
m
o
=;
v, = mg
k
(d) The sky diver's descent speed is initially V o and the acceleration is (close to) g. However,
once the parachute opens, the force of air resistance provides a large (speed-dependent)
upward acceleration, causing her descent velocity to decrease. The slope of the v vs. t graph
(the acceleration) is not constant but instead decreases to zero as her descent speed decreases
from va to vI' Therefore, the graph is not linear.
rng
Vt=T
time,t - -
440 • CRACKING THE AP PHYSICS EXAM
\I
(e) Since a = dvldt and, from part (b), a = (mg -kv) lm, we have
dv mg -kv
=
dt
m
-
~
dv
mg- kv
dt
m
Integrating both sides of this equation gives
- t lnlmg- kvl= -;);- t+ c
where c is a constant of integration. This equation can be rewritten in the form
mg - kv =Cel-klm)1
kv =mg-CeH l m)1
Since v
=Vo at t =0, we can determine the constant C:
kv o = mg - C
~
C=mg -kvo
The refore, the equation for the sky diver's descent speed as a function of time is
v(t) = t[mg-(mg -kvo)eH lm)l]
5. (a) The forces acting on a person standing against the cylinder wall are gravity (Fw' downward),
the normal force from the wall (FN, directed toward the center of the cylinder ), and the force of
static friction (Fp dire cted upward):
p assen ger
(mass = m)
(b) In order to keep the passenger from sliding down the wa ll, the maximum force of static
friction must be at least as gre at as the passenger's weight: F l lm x) ;:: mg. Since FI lm,,) = Jl ,FN' this
condition becomes
SOLUTIO NS TO THE CH APTER RE VIEW QUESTIO NS .
441
Now, consid er the circular motion of the passenger . Neither F,nor Fw has a component toward
the center of the path, so the centripetal force is provided entirely by the normal force:
mv 2
FN = - r
Substituting this expression for FN into the previous equation, we get
mo'
Jls- - ? mg
r
> gr
Jls - 2
v
Therefore, the coefficient of static friction between the passenger and the wall of the cylind er
must satisfy this condition in order to keep the pa ssenger from sliding down.
(c) Since th e mass m canceled out in deriving the expression for Jls' the conditions are
independent of mass. Thus, the inequality Jl,? gr/ v' holds for both the adult passenger of
mass m and the child of mass m/2.
6. (a) The forces actin g on the car are gravity (Fw' downward), the normal force from the road (FN'
upward), and the force of sta tic friction (Fe directed toward the center of curvatur e of th e roa d):
tow ard center
of curvature ¢:= F, +-- --ocar (mass = m)
of the road
(b) The force of static friction [we assume static friction because we don't want the car to slide
(that is, skid)] provides the necessary centripetal force:
mv 2
F, = - r
442 • CRACKI NG THEAP PHYSICS EXAM