Download AP-1 Cutnell 00-05 1st Sem Rev Key Points

Ex. 3: A tall building casts
a shadow that is 67.2 m
long. The angle between
the sun’s rays and the
ground is θ = 50.0°.
Determine the height of
the building.
Ex. 4: A lakefront drops off
gradually at an angle θ. At
14.0 m offshore the depth is
2.25 m.
(a) What is the value of θ ?
(b) What would be the depth
d of the lake at a distance of
22.0 m from the shore?
Ex. 7: A displacement
vector r has a magnitude
of r = 175 m and points at
an angle of 50.0° relative
to the x axis. Find the x
and y components
of this vector.
CH 2
Ex. 5 - A parachute and
brakes begin slowing a
drag racer at t0 = 9.0 s, the
velocity is v0 = +28 m/s.
When t = 12.0 s, the
velocity has been reduced
to v = +13 m/s. What is the
average acceleration of the
dragster?
Kinematic equations
v = v0 + at
x = vt = 1/2 (v0 + v)t
2
x = v0t + 1/2 at
2
2
v = v0 + 2ax
Ex. 6 - A speedboat has a
constant acceleration of
2
+2.0 m/s . If the initial
velocity of the boat is
+6.0 m/s, find its
displacement after
8.0 seconds.
Ex. 7 - A jet is taking off the
deck of an aircraft carrier.
Starting from rest, the jet is
catapulted with a constant
acceleration of +31 m/s2 along a
straight line and reaches a
velocity of +62 m/s. Find the
displacement of the jet.
Ex 13 - A stone is
dropped from rest
from the top of a tall
building. After 3.00 s of
freefall, what is the
displacement y of the
stone?
Ex 14 - After 3.00 s
of freefall, what is
the velocity v of
the stone in the
previous example?
The acceleration due
to gravity also affects
the motion of upward
moving objects,
decreasing their
velocity.
Ex 15 - A referee tosses a
coin with an initial speed
of 8.00 m/s. Neglecting air
resistance, how high does
the coin go above its point
of release?
Ex 16 - What
is the total
time the coin
is in the air?
The coin in the previous
example begins with an
upward velocity which
decreases to zero,
then increases in a
downward direction, but
the downward acceleration
due to gravity is constant.
Velocity and
acceleration can be
plotted on a graph as
displacement vs.
time and as velocity
vs. time.
In each case, the slope
of the line indicates the
velocity (rate of change
of displacement) and
acceleration (rate of
change of velocity).
If an object is
accelerating, a position
vs. time graph will be a
curved line. The velocity
at any instant in time is
the slope of a line tangent
to the curve at that point.
The slope of the
line of a velocity vs.
time graph is
the average
acceleration.
CH 3
In projectile motion, the
vx is constant (vx = v0x),
ax = 0.
But vy changes because
of the acceleration
due to gravity;
2
ay = 9.80 m/s usually.
Ex. 2 - An airplane flies horizontally
with a constant velocity of +115 m/s
at an altitude of 1050 m. The plane
releases a “care package” that falls
to the ground along a curved
trajectory. Ignoring air resistance,
determine the time required for the
package to hit the ground.
Ex. 3 - Find the
velocity of the
package just
before it hits the
ground.
Ex. 5 - A placekicker kicks a
football at an angle of
θ = 40.0° above the
horizontal axis. The initial
speed of the ball is v0 = 22
m/s. Find the maximum
height H that the ball attains.
Ex. 6 - Determine
the time of flight of
the football
between kickoff
and landing.
Ex. 7 Calculate the
range R of the
kickoff.
CH 4
Newton’s First Law objects resist
acceleration.
Law of inertia.
The mass of an object
is a quantitative
measure of inertia.
Newton’s second law F = ma.
The unit of force is the
Newton.
1 Newton =
2
1kg•m/s .
Newton’s Third Law - for
every action (force) there
is an equal and opposite
reaction. The third law
describes two different
forces being applied to
two different objects.
Free-body diagram
A diagram that
represents the object
and the forces that
act on it.
Newton’s Law of Universal
Gravitation - every particle in
the universe exerts a force on
every other particle by the
following equation:
F = G m1m2 /r2
G = 6.67259 x
-11
10
2
2
N•m /kg
Ex. 5 - What is the
magnitude of the
gravitational force that
acts on each particle,
assuming
m1 = 12 kg, m2 = 25 kg,
and r = 1.2 m?
Relation Between Mass
and Weight 2
W = G ME m2 /r
W=mg
g is determined by G, ME, and r.
The normal force FN is one
component of the force that
a surface exerts on an
object with which it is in
contact, namely, the
component that is
perpendicular to the
surface.
The forces on the block are
balanced. Adding another
downward force causes FN to
increase. Adding an upward force,
subtracts from W and FN decreases;
possibly to zero.
Ex. 8 - In a circus balancing act, a
woman performs a headstand on top
of a man’s head. The woman weighs
490 N, and the man’s head and neck
weigh 50 N. It is primarily the seventh
cervical vertebra in the spine that
supports all the weight above the
shoulders. What is the normal force
that this vertebra exerts on the neck
and head of the man (a) before the act
and (b) during the act?
Friction before an object
starts to move is static
friction. The maximum static
frictional force, fSMAX, is the
maximum frictional force that
can be applied before motion
and acceleration occur.
The ratio of fSMAX to FN
is called the coefficient
of static friction. It is
represented by µS.
fSMAX /FN= µS
fSMAX /FN = µS
This equation relates
only the magnitudes
of the vectors, not
the directions.
Ex. 8 - A sled is resting on a
horizontal patch of snow, and
the coefficient of static friction
is µS = 0.350. The sled and
its rider have a total mass of
38.0 kg. Determine the
horizontal force needed to
start the sled barely moving.
When two surfaces are moving
across each other, the friction is
called kinetic friction fK. It is
usually less than static friction.
µK is the coefficient of kinetic
friction.
fK/FN = µK
Ex. 10 - A sled is traveling
at 4.00 m/s along a
horizontal stretch of
snow. The coefficient
of kinetic friction is mK =
0.0500. How far does the
sled go before stopping?
Tension - the force on a rope
that would tend to pull the rope
apart. A force on one end of a
rope will be felt by the object to
which it is attached. We will
assume for our discussions that
all ropes are massless.
Equilibrium - An object is in
equilibrium when it has zero
acceleration. The x-component
of the net force and the ycomponent of the net force must
both be zero.
∑Fx = 0
∑Fy = 0
If an object is accelerating,
these new equations apply:
∑Fx = max
∑Fy = may
Ex. 16 - An 8500-kg truck is hauling
a 27 000-kg trailer along
a level road. The acceleration is
0.78 m/s2. Ignoring the retarding
forces of friction and air resistance,
determine (A) the magnitude of the
tension in the horizontal drawbar
between the trailer and the truck
and (B) the force D that propels the
truck forward.
Ex. 18 - A flatbed truck is
carrying a crate up a 10.00° hill.
The coefficient of static friction
between the truck bed and the
crate is µs = 0.350. Find the
maximum acceleration that the
truck can attain before the crate
begins to slip backward relative
to the truck.
Ex. 19 - Block 1 (m1 = 8.00 kg)
is moving on a frictionless 30.0°
incline. This block is connected
to block 2 (m2 = 22.0 kg) by a
cord that passes over a
massless and frictionless pulley.
Find the acceleration of each
block and the tension in the
cord.
CH 5
In uniform circular motion the
magnitude of the velocity
vector is constant. The
direction; however, is
constantly changing. This
change is an acceleration
which is called
“centripetal acceleration”.
ac =
2
v /r
The direction of
the vector
quantity ac is
toward the center
of the circle.
Ex. 3 - The bobsled track
at the Olympics contained
turns with radii of 33 m and
24 m. Find the centripetal
acceleration for each turn for
a speed of 34 m/s.
Express the answers as
2
multiples of g = 9.8 m/s .
An object in uniform circular
motion is constantly being
accelerated. This is a noninertial frame of reference;
and an object in uniform
circular motion can never be
in equilibrium.
The force that pulls an
object into a circular path
is called a centripetal
force. F = ma, so Fc =
mac; therefore,
Fc =
2
mv /r
Ex. 5 - A model airplane has
a mass of 0.90 kg and moves
at a constant speed on a
circle that is parallel
to the ground. Find the
tension T in the guideline
(length = 17 m) for speeds of
19 and 38 m/s.
In vertical circular motion the
centripetal force is the vector sum
of the normal force and the
component of the weight that
pushes directly toward the center of
the circle. In the lower half of the
circular motion the centripetal force
will be less (normal force minus
weight) than in the upper half of the
circle (normal force plus weight).
A the top of the circle, the
centripetal force is normal force
plus weight:
2
FC = mv /r = FN + mg
At the correct speed, normal force
can become zero and:
FC = mv2/r = mg
Solving the last two terms for v
gives:
v= √rg
At this speed, the track
does not exert a normal
force; mg provides all
the centripetal force.
The rider at this point
experiences apparent
weightlessness.
Ex. 15 - A roller coaster
loop has a radius of 10
meters. What is the
minimum velocity
required to keep the
cars in the loop during
the ride?
Ex. 16 - An evil father is
pushing his daughter in a
swing. If he gleefully pushes
her as hard as he can, what
is the minimum velocity at
which she will make a
complete vertical circle on
the swingset if the swing’s
chain is 6 meters long?
Ex. 17 - After a lengthy trial, the
court decides that the punishment
should fit the crime. The father is
sentenced to be pushed in a circle
on the same swingset. If he weighs
six times as much as his daughter,
what is his minimum speed to
complete the circle? What other
differences are there in the
execution of his punishment?