Download Version PREVIEW – Semester 1 Review – Slade – (22222) 1 This

Version PREVIEW – Semester 1 Review – Slade – (22222)
This print-out should have 48 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Holt SF 02Rev 10A
001 (part 1 of 2) 10.0 points
Consider the position-time graph for a squirrel running along a clothesline.
b
position (m)
4
vavg =
1
∆x
3.0 m
=
= 2 m/s .
∆t
1.5 s
Holt SF 02A 02
003 10.0 points
If Joe rides south on his bicycle in a straight
line for 25 min with an average speed of 12.6
km/h, how far has he ridden?
Correct answer: 5.25 km.
3
Explanation:
b
b
2
1
Let :
0 b
1
−1
2
3
4b
∆t = 25 min and
vavg = 12.6 km/h .
5
b
−2
time (s)
What is the squirrel’s displacement at 1.5
s?
Your answer must be within ± 5.0%
Correct answer: 3 m.
∆x = vavg ∆t
= (12.6 km/h)(25 min) ·
1h
60 min
= 5.25 km
toward the south.
Explanation:
Let :
xi = 0 m .
The final position is
2 m + (4 m)
x1 + x2
=
2
2
= 3.0 m
xf =
and the displacement is
∆x = xf − xi = 3.0 m − (0 m)
= 3.0 m .
002 (part 2 of 2) 10.0 points
What is the squirrel’s average velocity during
the time interval between 0.0 s and 1.5 s?
Your answer must be within ± 5.0%
Correct answer: 2 m/s.
Explanation:
Let :
Velocity Relationships 01
004 10.0 points
Consider three position curves between time
points tA and tB .
s
sA A
3
2
1
sB
B
0 tA
tB t
Choose the correct relationship among
vA + vB
quantities v 1 , v 2 , and v3 . v =
when
2
a is a constant.
1. v 1 = v 2 = v 3 correct
2. v 1 > v 2 > v 3
3. v 1 < v 2 < v 3
∆t = 1.5 s .
Explanation:
Version PREVIEW – Semester 1 Review – Slade – (22222)
2
The average velocity of an object is
007 (part 3 of 4) 10.0 points
Find the instantaneous velocity at 4.5 s.
sB − sA
displacement
=
.
time
tB − tA
v=
All three curves have exactly the same
change in position ∆s = sB − sA in exactly
the same time interval ∆t = tB − tA , so all
three average velocities are equal:
Correct answer: 0 m/s.
Explanation:
v=
v1 = v2 = v3 .
008 (part 4 of 4) 10.0 points
Find the instantaneous velocity at 8 s.
Serway CP 02 15
005 (part 1 of 4) 10.0 points
The position versus time for a certain object
moving along the x-axis is shown. The object’s initial position is −4 m.
10
Correct answer: 2 m/s.
Explanation:
b
position (m)
8
v=
6
b
4
b
2
b
0
−2
b
−4
−6
0
b
1
2
3
4 5
time (s)
6
7
8
9
Find the instantaneous velocity at 0.5 s.
Correct answer: 13 m/s.
Explanation:
The instantaneous velocity is the slope of
the tangent line at that point.
v=
9 m − (−4 m)
= 13 m/s .
1s−0s
Explanation:
v=
4 m − (9 m)
= −2.5 m/s .
3s−1 s
0 m − (−4 m)
= 2 m/s .
9s−7s
Holt SF 02Rev 54
009 10.0 points
A tennis ball with a velocity of 11.3 m/s to
the right is thrown perpendicularly at a wall.
After striking the wall, the ball rebounds in
the opposite direction with a velocity of −8.50
m/s to the left.
If the ball is in contact with the wall for
0.011 s, what is the average acceleration of
the ball while it is in contact with the wall?
Correct answer: −1800 m/s2 .
Explanation:
Let : vi = +11.3 m/s ,
vf = −8.50 m/s ,
∆t = 0.011 s .
006 (part 2 of 4) 10.0 points
Find the instantaneous velocity at 2 s.
Correct answer: −2.5 m/s.
4 m − (4 m)
= 0 m/s .
6s−3s
ā =
and
vf − vi
−8.5 m/s − 11.3 m/s
=
∆t
0.011 s
= −1800 m/s2 .
Holt SF 02B 03
Version PREVIEW – Semester 1 Review – Slade – (22222)
010 10.0 points
With an average acceleration of −1.5 m/s2 ,
how long will it take a cyclist to bring a bicycle
with an initial speed of 14.5 m/s to a complete
stop?
Let : v0 = 10.1 m/s and
g = 9.8 m/s2 .
Basic Concept:
For constant acceleration, we have
Correct answer: 9.66667 s.
Explanation:
Let :
3
v = v0 + a t .
aavg = −1.5 m/s2 ,
vi = 14.5 m/s , and
vf = 0 m/s .
(1)
Solution: The velocity at the top is zero.
Since we know velocities and acceleration, Eq.
1 containing v, a, and t . Choose the positive
direction to be up; then a = −g and
0 = v0 + (−g) tup
Ball N 03
011 10.0 points
A ball is thrown upward. Its initial vertical speed is 10.1 m/s , acceleration of gravity
is 9.8 m/s2 , and maximum height hmax are
shown in the figure below.
Neglect: Air resistance. The acceleration
of gravity is 9.8 m/s2 .
b
b
b
b
b
b
b
b
b
9.8 m/s2
b
hmax
10.1 m/s
bbb
bb
b b
b b
b
What is its time interval, tup , between the
release of the ball and the time it reaches its
maximum height?
or
v0
g
(10.1 m/s)
=
(9.8 m/s2 )
= 1.03061 s .
tup =
Ball Thrown Up 12
012 10.0 points
A ball is thrown straight up and passes point
B (at a height of 48.8 m above its starting
point O) in 5.2 s.
The acceleration of gravity is 9.8 m/s2 .
tA is the time
b b b
hA
b b
b
b
b
to reach its
A
b
b
y
b
b
maximum
b
b
height hA
b
b B
48.8 m
vf − vi
∆v
vi
=
=−
∆t
∆t
∆t
−14.5 m/s
−vi
=
= 9.66667 s .
∆t =
a
−1.5 m/s2
a=
b
b
b
v0
b
b
b
b
b
O 5.2 s
tA
Figure is not drawn to scale.
What was its initial speed k~v0 k ?
Correct answer: 34.8646 m/s.
Explanation:
Basic Concept: Motion under uniform
acceleration
Correct answer: 1.03061 s.
Explanation:
t
y = v0 t +
1 2
at .
2
Version PREVIEW – Semester 1 Review – Slade – (22222)
Solution: The initial velocity is directed
upward and gravity acts downward, so
y = v0 t −
1 2
gt ,
2
Basic Concepts:
Horizontally,
∆x = vx ∆t
and
1 2
g t , and
2
y + 21 g t2
v0 =
t
48.8 m + 21 (9.8 m/s2 ) (5.2 s)2
=
5.2 s
= 34.8646 m/s .
v0 t = y +
Drop a Rock
013 10.0 points
If you drop a rock from a height of 12 m , it accelerates at g and strikes the ground 1.56492 s
later.
The acceleration of gravity is 9.8 m/s2 .
If you drop the same rock from half that
height, its acceleration will be
since ax = 0 m/s2 .
Vertically,
1
∆y = − g (∆t)2
2
since vi,y = 0 m/s.
Given:
∆y = −0.809 m
∆x = 18.3 m
g = 9.81 m/s2
Solution:
From the horizontal motion,
∆t =
1. more.
2. 0.
3. unable to determine.
4. about half.
5. the same. correct
Explanation:
The acceleration of an object due to gravity
is virtually constant at the earth’s surface and
is not affected by height.
4
∆x
vx
so that
2
1
∆x
∆y = − g
2
vx
s
−g(∆x)2
vx =
2∆y
s
− (9.81 m/s2 ) (18.3 m)2
=
2(−0.809 m)
= 45.0605 m/s .
Holt SF 03Rev 34
014 10.0 points
The fastest recorded pitch in Major League
Baseball was thrown by Nolan Ryan in 1974.
If this pitch were thrown horizontally, the ball
would fall 0.809 m (2.65 ft) by the time it
reached home plate, 18.3 m (60 ft) away.
The acceleration of gravity is 9.81 m/s2 .
How fast was Ryan’s pitch?
Accelerating Elevator
015 10.0 points
An elevator starts from rest with a constant
upward acceleration and moves 1 m in the first
1.9 s. A passenger in the elevator is holding a
4.8 kg bundle at the end of a vertical cord.
The acceleration of gravity is 9.8 m/s2 .
What is the tension in the cord as the elevator accelerates?
Correct answer: 45.0605 m/s.
Correct answer: 49.6993 N.
Explanation:
Explanation:
Version PREVIEW – Semester 1 Review – Slade – (22222)
v 2 − vo2
−vo2
=
.
2 ∆x
2 ∆x
a=
T
g
aelevator
5
X
~ = m~a to express the force exApply
F
erted on the bullet by the wood:
mg
Let h be the distance traveled and a the
acceleration of the elevator. With the initial
velocity being zero, we simplify the following
expression and solve for acceleration of the
elevator:
1
1
h = v0 t + a t2 = a t2
2
2
2h
=⇒ a = 2 .
t
The equation describing the forces acting on
the bundle is
Fnet = m a = T − m g
T = m (g + a)
2h
=m g+ 2
t
2 (1 m)
2
= (4.8 kg) 9.8 m/s +
(1.9 s)2
= 49.6993 N .
Tipler PSE5 04 29
016 10.0 points
A bullet of mass 0.0024 kg moving at 550 m/s
impacts a large fixed block of wood and travels
6.8 cm before coming to rest.
Assuming that the acceleration of the bullet
is constant, find the force exerted by the wood
on the bullet.
Correct answer: −5.33824 kN.
Fwood = m a
m vo2
=−
2 ∆x
1 kN
(0.0024 kg) (550 m/s)2
×
=−
2 (0.068 m)
1000 N
= −5.33824 kN .
Jumping From the Fire
017 10.0 points
A person of mass 73.4 kg escapes from a burning building by jumping from a window situated 39.4 m above a catching net.
The acceleration of gravity is 9.8 m/s2 .
If air resistance exerts a force of 106.2 N
on him as he falls, determine his speed just
before he hits the net.
Correct answer: 25.6559 m/s.
Explanation:
The forces acting on him are the gravitational force m g acting downward, and the air
resistance, Fa , acting upward. The acceleration is downward and the net force is
Fnet = m a = m g − Fa
a=
The person is in free fall, so his final speed is
defined by
Explanation:
m = 0.0024 kg ,
vo = 550 m/s , and
∆x = 6.8 cm = 0.068 m .
The deceleration of the bullet is constant, so
v 2 = vo2 + 2 a ∆x = 0
√
2ah
r
2 (m g − Fa ) h
=
m
s
vf =
Let :
m g − Fa
m
=
2 [(73.4 kg)(9.8 m/s2 ) − 106.2 N] (39.4 m)
73.4 kg
= 25.6559 m/s.
Version PREVIEW – Semester 1 Review – Slade – (22222)
6
3. No; it is balanced.
Hewitt CP9 04 E11
018 10.0 points
In the orbiting space shuttle you are handed
four identical boxes. The first one is filled
with sand. The second one is filled with iron.
The third one is filled with water. The last
one is filled with feathers. Shake the boxes.
Which one offers the greatest resistance and
which one offers the smallest resistance?
1. iron, water
2. sand, water
3. iron, feathers correct
4. Yes; upward.
Explanation:
The horizontal components of the two 30 N
forces cancel, leaving an upward force that is
less than 60 N. Thus, the net force on the box
is down, causing it to accelerate downward.
Conceptual 05 Q8
020 10.0 points
The Earth exerts an 760 N gravitational force
on a man.
What force does the man exert on the
Earth?
Correct answer: 760 N.
4. feathers, iron
Explanation:
By Newton’s third law, the man exerts an
equal but opposite force on the Earth.
5. All are wrong.
Explanation:
Among these three materials, iron has the
largest density and feathers have the smallest
density. So the box filled with iron has largest
mass and offers greatest resistance while the
box filled with feathers has the smallest mass
and offers the smallest resistance.
Conceptual 04 Q17
019 10.0 points
Two 30 N forces and a 60 N force act on a
hanging box as shown.
30 N
30 N
Hewitt CP9 05 E33
021 10.0 points
Consider a stone at rest on the ground. There
are two interactions that involve the stone.
One is between the stone and the Earth; Earth
pulls down on the stone and the stone pulls
up on the Earth.
What is the other interaction?
1. between the ground and air
2. between the stone and the ground correct
3. between the ground and the Earth
4. All are wrong.
5. between the Earth and air
60 N
Will the box experience acceleration?
1. Unable to determine without the angle.
2. Yes; downward. correct
Explanation:
If the action is the stone pushing down on
the ground surface, the reaction is the ground
pushing up on the stone. This upward force
on the stone is called the normal force.
Force and Motion 15
022 (part 1 of 2) 10.0 points
Version PREVIEW – Semester 1 Review – Slade – (22222)
The following 2 questions refer to the collisions between a car and a truck whose weight
is much heavier than the car (M ≫ m). For
each description of a collision below, choose
the one answer from the possibilities that best
describes the size (or magnitude) of the forces
between the car and the truck.
Assume: Friction is so small that it can be
ignored.
M
v
m
v
They are both moving at the same speed
when they collide.
1. Neither exerts a force on the other; the
car gets smashed simply because it is in the
way of the truck.
2. The truck exerts a greater amount of force
on the car than the car exerts on the truck.
3. Not enough information is given to pick
one of these answers.
4. The truck exerts the same amount of
force on the car as the car exerts on the truck.
correct
5. None of these answers describes the situation correctly.
6. The car exerts a greater amount of force
on the truck than the truck exerts on the
car.
Explanation:
By Newton’s third law, action and reaction
are of the same magnitude and in the opposite
direction.
7
on the car than the car exerts on the truck.
2. The truck exerts the same amount of
force on the car as the car exerts on the truck.
correct
3. None of these answers describes the situation correctly.
4. The car exerts a greater amount of force
on the truck than the truck exerts on the
car.
5. Neither exerts a force on the other; the
car gets smashed simply because it is in the
way of the truck.
6. Not enough information is given to pick
one of these answers.
Explanation:
The same reason as Part 1.
Holt SF 04Rev 63
024 (part 1 of 5) 10.0 points
Three blocks are in contact with each other
on a frictionless horizontal surface. A 335 N
horizontal force is applied to the block with
mass of 2.9 kg as shown in the figure below.
The acceleration of gravity is 9.8 m/s2 .
F
2.9 kg
5.7 kg
7.5 kg
a) What is the net force on the block with
mass 2.9 kg?
Correct answer: 60.3416 N.
Explanation:
023 (part 2 of 2) 10.0 points
The car is moving much faster than the heavier truck when they collide.
1. The truck exerts a greater amount of force
F
m1
m2
m3
Version PREVIEW – Semester 1 Review – Slade – (22222)
Given :
F
m1
m2
m3
= 335 N ,
= 2.9 kg ,
= 5.7 kg ,
= 7.5 kg ,
8
026 (part 3 of 5) 10.0 points
c) What is the resultant force on the block
with mass 7.5 kg?
and
Correct answer: 156.056 N.
2
g = 9.8 m/s .
Explanation:
Solution:
Basic Concepts:
F3,net = m3 a
Fnet = m a
= (7.5 kg) 20.8075 m/s2
= 156.056 N
mtotal = m1 + m2 + m3
Solution:
N
F
m1 + m2 + m3
g
(m1 + m2 + m3 ) g
027 (part 4 of 5) 10.0 points
d) What is the magnitude of the force between
the block with mass 5.7 kg and 7.5 kg?
Correct answer: 156.056 N.
Explanation:
a
F = (m1 + m2 + m3 ) a .
F2,3
The acceleration of the system is
m3
F
a=
m1 + m2 + m3
335 N
=
2.9 kg + 5.7 kg + 7.5 kg
= 20.8075 m/s2
m3 g
Basic Concept:
F3,net = m3 a = F2,3
Solution:
to the right.
F2,3 = m3 a
F1,net = m1 a
= (2.9 kg) 20.8075 m/s2
= 60.3416 N
025 (part 2 of 5) 10.0 points
b) What is the resultant force on the block
with mass 5.7 kg?
= (7.5 kg) 20.8075 m/s2
= 156.056 N
028 (part 5 of 5) 10.0 points
e) What is the magnitude of the force between
the block with mass 2.9 kg and 5.7 kg?
Correct answer: 274.658 N.
Correct answer: 118.602 N.
Explanation:
a
Explanation:
Solution:
F1,2
= (5.7 kg) 20.8075 m/s2
= 118.602 N
F2,3
m2
F2,net = m2 a
m2 g
Version PREVIEW – Semester 1 Review – Slade – (22222)
Basic Concept:
F2,net = m2 a = F1,2 − F2,3
Solution: Since F2,3 = 156.056 N is provided
in the previous part, we have
F1,2 = m2 a + F2,3
= (5.7 kg) 20.8075 m/s2 + 156.056 N
= 274.658 N
Down a Smooth Incline
029 (part 1 of 3) 10.0 points
A 3.98 kg block slides down a smooth, frictionless plane having an inclination of 24◦ .
The acceleration of gravity is 9.8 m/s2 .
9
Motion has a constant acceleration. Recall
the kinematics of motion with constant acceleration.
Solution: Because the block slides down
along the plane of the ramp, it seems logical
to choose the x-axis in this direction. Then
the y-axis must emerge perpendicular to the
ramp, as shown.
Let us now examine the forces in the xdirection. Only the weight has a component
along that axis. So, by Newton’s second law,
X
Fx = m a = m g sin θ
Thus
a = g sin θ
With our particular value of θ,
a = (9.8 m/s2 ) sin 24◦ = 3.98602 m/s2
2. 8
8m
8
3. 9
kg
24◦
Find the acceleration of the block.
Correct answer: 3.98602 m/s2 .
m = 3.98 kg
θ = 24◦ .
and
Consider the free body diagram for the
block
N
θ
sin
g
m
N
=
Correct answer: 4.79161 m/s.
Explanation:
Since v0 = 0, a = 3.98602 m/s2 and L =
2.88 m,
Explanation:
Given :
030 (part 2 of 3) 10.0 points
What is the block’s speed when, starting from
rest, it has traveled a distance of 2.88 m along
the incline.
θ
cos
g
m
W = mg
Basic Concepts:
Fx,net = F cos θ − W|| = 0
W|| = m g sin θ = m a
vf2 = v02 + 2 a (x − x0 )
= 2 (3.98602 m/s2 ) (2.88 m)
vf = 4.79161 m/s .
It is very important to note at this point that
neither of these values depended on the mass
of the block. This may seem odd at first, but
recall what Galileo discovered 300 years ago –
objects of differing mass fall at the same rate.
031 (part 3 of 3) 10.0 points
What is the magnitude of the perpendicular
force that the block exerts on the surface of
the plane at a distance of 2.88 m down the
incline?
Correct answer: 35.6319 N.
Explanation:
| FN | = m g cos θ
= 3.98 kg (9.8 m/s2 ) cos 24◦
= 35.6319 N.
Atwood Machine 15
032 10.0 points
A light, inextensible cord passes over a light,
frictionless pulley with a radius of 9.5 cm.
It has a(n) 16 kg mass on the left and a(n)
7.6 kg mass on the right, both hanging freely.
Initially their center of masses are a vertical
distance 1.2 m apart.
The acceleration of gravity is 9.8 m/s2 .
9.5 cm
ω
7.6 kg
a
T
10
m1 g
16 kg
m2 g
=⇒
a
By examining the free body diagram again,
we see that the force in the y direction is given
by
X
F = −m g cos θ = FN .
T
Version PREVIEW – Semester 1 Review – Slade – (22222)
Since the larger mass will move down and
the smaller mass up, we can take motion
downward as positive for m2 and motion upward as positive for m1 . Apply Newton’s
second law to m1 and m2 respectively and
then combine the results:
For mass 1:
X
F1 : T − m1 g = m1 a
(1)
For mass 2:
X
F2 :
m2 g − T = m2 a
(2)
We can add Eqs. (1) and (2) above and obtain:
m2 g − m1 g = m1 a + m2 a
m2 − m1
g
m1 + m2
16 kg − 7.6 kg
=
(9.8 m/s2 )
16 kg + 7.6 kg
= 3.48814 m/s2 .
a=
16 kg
1.2 m
7.6 kg
At what rate are the two masses accelerating when they pass each other?
Correct answer: 3.48814 m/s2 .
T = m1 (g + a)
= (7.6 kg) (9.8 m/s2 + 3.48814 m/s2 )
= 100.99 N .
Explanation:
Let : R = 9.5 cm ,
m1 = 7.6 kg ,
m2 = 16 kg ,
h = 1.2 m , and
v = ωR.
Consider the free body diagrams
Centripetal Acceleration
033 10.0 points
A car rounds a curve while maintaining a
constant speed.
Is there a net force on the car as it rounds
the curve?
1. It depends on the sharpness of the curve
and speed of the car.
Version PREVIEW – Semester 1 Review – Slade – (22222)
2. No – its speed is constant.
3. Yes. correct
Explanation:
Acceleration is a change in the speed and/or
direction of an object. Thus, because its
direction has changed, the car has accelerated
and a force must have been exerted on it.
11
Suppose an automobile has a kinetic energy
of 2500 J.
When it moves with five times the speed,
what will be its kinetic energy?
Correct answer: 62500 J.
Explanation:
Let
Eold = 2500 J
n = 5.
and
Horizontal Circle 04
034 10.0 points
A 3.08 kg mass attached to a light string
rotates on a horizontal, frictionless table. The
radius of the circle is 0.691 m, and the string
can support a mass of 30 kg before breaking.
The acceleration of gravity is 9.8 m/s2 .
What maximum speed can the mass have
before the string breaks?
The kinetic energy K ∝ v 2 , so for
vnew = n · vold ,
Correct answer: 8.12152 m/s.
and
Explanation:
The string will break when the tension exceeds the weight corresponding to 25 kg, so
Tmax = M g = (30 kg)(9.8 m/s2 ) .
When the smaller mass rotates in a horizontal
circle, the string tension provides the centripetal force, so
m v2
r
r
rT
.
v=
m
Thus the maximum velocity before the string
breaks is
r
r Tmax
vmax =
m
s
(0.691 m)(30 kg)(9.8 m/s2 )
=
3.08 kg
T =
Knew
(n · vold )2
v2
=
= n2
= new
2
2
Kold
vold
vold
Knew = n2 Kold
= (5)2 (2500 J)
= 62500 J .
Dragging a Block 02
036 (part 1 of 5) 10.0 points
An 18.1 kg block is dragged over a rough,
horizontal surface by a constant force of 178 N
acting at an angle of angle 28.1 ◦ above the
horizontal. The block is displaced 96.2 m,
and the coefficient of kinetic friction is 0.211.
178
N
18.1 kg
µ = 0.211
Find the work done by the 178 N force. The
acceleration of gravity is 9.8 m/s2 .
= 8.12152 m/s .
Correct answer: 15105.2 J.
Kinetic Energy Comparison
035 10.0 points
◦
28. 1
Explanation:
Consider the force diagram
Version PREVIEW – Semester 1 Review – Slade – (22222)
12
What is the sign of the work done by the
frictional force?
n
F
θ
fk
1. zero
2. positive
3. negative correct
mg
Explanation:
~ =F
~ · ~s, where ~s is the distance
Work is W
traveled. In this problem ~s = 5 x̂ is only in
the x̂ direction.
WF = Fx sx = F cos θ sx
= (178 N) cos 28.1◦ (96.2 m)
= 15105.2 J .
037 (part 2 of 5) 10.0 points
Find the magnitude of the work done by the
force of friction.
Correct answer: 1898.69 J.
Explanation:
To find the frictional force, Ff riction = µ N ,
we need to find N from vertical force balance.
N is in the same direction as the y component
of F and opposite the force of gravity, so
F sin θ + N = m g
and
N = m g − F sin θ .
039 (part 4 of 5) 10.0 points
Find the work done by the normal force.
Correct answer: 0 J.
Explanation:
Since the normal force points in the ŷ direction and ~s is in the x̂ direction
~ · ~s = N ŷ · x̂ = 0 .
WN = N
040 (part 5 of 5) 10.0 points
What is the net work done on the block?
Correct answer: 13206.5 J.
Explanation:
The net work done on the body is the algebraic sum of the work done by the external
force F and the work done by the frictional
force
Wnet = WF + Wµ
= 15105.2 J + −1898.69 J
= 13206.5 J .
Thus the friction force is
~ f riction = −µ N x̂ = −µ (m g − F sin θ x̂
F
and the work done by friction is
~ f riction ~s = −|Ff | |s|
Wµ = F
= −µ (m g − F sin θ) sx
= −(0.211)[(18.1 kg)(9.8 m/s2 )
−(178 N) sin 28.1◦ ](96.2 m)
= −1898.69 J
|Wµ | = 1898.69 J .
038 (part 3 of 5) 10.0 points
Holt SF 06Rev 55
041 (part 1 of 2) 10.0 points
A constant force of 3.0 N to the right acts on
a 1.4 kg mass for 0.48 s.
a) Find the final velocity of the mass if it is
initially at rest.
Correct answer: 1.02857 m/s.
Explanation:
Let to the right be positive.
Let : F = 3.0 N ,
m = 1.4 kg ,
t = 0.48 s .
and
Version PREVIEW – Semester 1 Review – Slade – (22222)
Since vi = 0 m/s,
~ ∆t = m ∆~v = m~vf − m~vi = m~vf
F
F ∆t
m
(3 N) (0.48 s)
=
1.4 kg
vf =
= 1.02857 m/s
to the right.
042 (part 2 of 2) 10.0 points
b) Find the final velocity of the mass if it
is initially moving along the x-axis with a
velocity of 3.9 m/s to the left.
Correct answer: −2.87143 m/s.
Explanation:
Let : vi = −3.9 m/s
~ ∆t = m ∆~v = m~vf − m~vi
F
F ∆t + m vi
vf =
m
F ∆t
+ vi
=
m
(3 N) (0.48 s)
=
+ (−3.9 m/s)
1.4 kg
= −2.87143 m/s ,
Applying impulse,
~ ∆t = m~vf − m~vi = m~vf
F
since vi = 0 m/s.
m vf
∆t
(0.57 kg) (9 m/s)
=
0.22 s
= 23.3182 N .
F =
Spring Between Blocks 02
044 10.0 points
Two blocks of masses M and 3 M are placed
on a horizontal, frictionless surface. A light
spring is attached to one of them, and the
blocks are pushed together with the spring
between them.
Correct answer: 23.3182 N.
3M
M
Before
(a)
v
3M
M
After
(b)
which is 2.87143 m/s to the left.
Serway CP 04 02
043 10.0 points
A football punter accelerates a 0.57 kg football from rest to a speed of 9 m/s in 0.22 s.
What constant force does the punter exert
on the ball?
13
A cord holding them together is burned,
after which the block of mass 3 M moves to
the right with a speed of 49 m/s. What is the
speed of the block of mass M?
Correct answer: 147 m/s.
Explanation:
Explanation:
Given : m = 0.57 kg ,
vf = 9 m/s, and
∆t = 0.22 s .
Let : m1 = M ,
m2 = 3 M ,
v2 = 49 m/s .
and
Version PREVIEW – Semester 1 Review – Slade – (22222)
14
From conservation of momentum ∆p = 0,
so
0 = m1 v1 + m2 v2
m2 v2
v1 = −
m1
(3 M ) (49 m/s)
=
M
= 147 m/s .
Pitching Machine Recoil
045 10.0 points
A baseball player uses a pitching machine to
help him improve his batting average. He
places the 66.1 kg machine on a frozen pond.
The machine fires a 0.108 kg baseball horizontally at a speed of 31.9 m/s.
What is the magnitude of the recoil velocity
of the machine?
Correct answer: 0.052121 m/s.
Explanation:
Blocks Compress a Spring 02
046 10.0 points
Two blocks of masses 4 kg and 15 kg are
placed on a horizontal, frictionless surface. A
light spring is attached to one of them, and
the blocks are pushed together with the spring
between them. A cord holding them together
is burned, after which the block of mass 15 kg
moves to the right with a speed of 27 m/s.
What is the velocity of the other mass in
m/s?
Correct answer: −101.25 m/s.
Explanation:
By momentum conservation
m1 v1 + m2 v2 = 0
so the velocity of m1 is
Let : m1 = 0.108 kg ,
m2 = 66.1 kg , and
v1f = 31.9 m/s .
We take the system to consist of the baseball and the pitching machine. Because of
the force of gravity and the normal force, the
system is not really isolated. However, both
of these forces are directed perpendicularly to
the motion of the system. Therefore, momentum is constant in the x direction because
there are no external forces in this direction
(assuming the surface is frictionless). The total momentum of the system before firing is
zero. Therefore, the total momentum after
firing must be zero, that is,
v1 = −
m2 v2
m1
Serway CP 04 15
047 (part 1 of 2) 10.0 points
Consider a 620 N cat burglar supported by a
cable as in the figure.
36.1◦
m1 v1f + m2 v2f = 0 ,
or, in components,
m1 v1f − m2 v2f = 0 .
v2f =
m1
0.108 kg
v1f =
(31.9 m/s )
m2
66.1 kg
= 0.052121 m/s .
Find the tension in the inclined cable.
Correct answer: 1052.28 N.
Explanation:
Version PREVIEW – Semester 1 Review – Slade – (22222)
Let : W = 620 N and
θ = 36.1◦ .
y
T2
θ
T1
x
W
Since the burglar is held in equilibrium,
X
Fy = 0
T2 sin θ − W = 0
620 N
W
=
sin θ
sin 36.1◦
= 1052.28 N .
T2 =
048 (part 2 of 2) 10.0 points
Find the tension in the horizontal cable.
Correct answer: 850.232 N.
Explanation:
Since the burglar is held in equilibrium,
X
Fx = 0
T1 − T2 cos θ = 0
T1 = T2 cos θ
= (1052.28 N) cos 36.1◦
= 850.232 N .
15