Version 097 – Exam 2 – hoffmann – (57870)
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Multiple-choice questions may continue on
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before answering.
001 (part 1 of 2) 10.0 points
A ball of mass m is attached to a spring of
negligible mass, and force constant k. The
spring is freely pivoted at the opposite end
from the ball, and the system is initially at
rest and horizontal.
L
L + ∆L
If the ball is allowed to fall, what is its
speed when the system is swinging through
the vertical and the spring is stretched an
L
amount ∆L =
, where L is the spring’s
10
initial, unstretched length?
s
k L2
1.
mg
s
kL
L
10 g −
2.
m
100
r
mgL
3.
10
s
L
kL
4.
correct
22 g −
10 m 10
s
kL
− 10 g L
5.
m
s
kL
6.
L
g−
m
s
kL
L
− 22 g
7.
10 m
10
8. Not enough information is given.
9.
s
kL
− 11 g
m
1
L
10
Explanation:
Using conservation of energy with E = K +
Us + Ug and taking Ug = 0 at the instant the
system is vertical we get
Ei = Ef
Ki + Usi + Ugi = Kf + Usf + Ugf
1
1
0 + 0 + m g (L + ∆L) = m v 2 + k (∆L)2
2 2
1
1
L2
11 L
2
= mv + k
mg
10
2
2
100
2
22 m g L k L
−
= m v2
10
100
mL
kL
22 g −
= mv 2
10
10 m
s
L
kL
=v
22 g −
10 m 10
002 (part 2 of 2) 10.0 points
Having calculated the speed in the first part,
now find the acceleration a of the ball at the
point where it swings through the vertical
position.
kL
− 2g
1. ĵ
110 m
kL
2. ĵ
− g correct
10 m
!
v2
3. ĵ g − 11 L
10
4. ĵ g
kL
+g
5. ĵ
10 m
kL
6. ĵ
10 m
7. 0
kL
8. ĵ 2g −
110 m
Version 097 – Exam 2 – hoffmann – (57870)
kL
9. −ĵ
−g
10 m
Explanation:
From Newton’s 2nd Law,
X
ma =
Fi = ĵ (k ∆L) − ĵ (m g)
kL
a = ĵ
−g .
10 m
2
Note that substituting v from part 1 into
v2
v2
a=
=
will generate a completely
r
L + ∆L
wrong answer, since the mass m is neither
moving in a circle nor at a constant speed.
003 10.0 points
A 3.02 kg ball is attached a fish line rated to
withstand 26.8 N. The ball is released from
rest at the horizontal position (θ0 = 90◦ ).
At what angle θ (measured from the vertical) will the fish line break? The acceleration
of gravity is 9.8 m/s2 .
1. 86.572
2. 67.2086
3. 79.8071
4. 61.8546
5. 58.2221
6. 72.4317
7. 75.3872
8. 59.3961
9. 78.1332
10. 82.483
Correct answer: 72.4317◦.
Explanation:
In the swing down to the breaking point,
energy is conserved:
m v2
2
m v2
.
2 g m cos θ =
r
At the breaking point, consider the radial
forces:
X
Fr = m ar
m g r cos θ =
m v2
T − m g cos θ =
r
2
T = 3 m g cos θ
T
θ = arccos
3mg
26.8 N
= arccos
3(3.02 kg)(9.8 m/s2 )
= 72.4317◦ .
004 10.0 points
A civil engineer is to design a road curve at a
95.2 m radius to carry racing cars traveling at
24 m/s.
µ=
0. 1
4
θ
At what angle θ should it be banked for
maximum control of the racing car? The
coefficient of friction of the surface is 0.14 and
the acceleration due to gravity is 9.8 m/s2 .
1. 45.363
2. 27.6897
3. 31.6908
4. 37.1672
5. 36.0329
6. 18.4982
7. 25.3378
8. 13.8944
9. 17.9385
10. 21.6596
Correct answer: 31.6908◦ .
Explanation:
Let :
v = 24 m/s ,
r = 95.2 m , and
g = 9.8 m/s2 .
The car will have optimum control when
the frictional force is zero, so the forces acting
Version 097 – Exam 2 – hoffmann – (57870)
on the car are the normal force and the force
due to gravity:
N cos θ
x
N sin θ
mg
To keep an object moving in a circle requires a force directed toward the center of
the circle; its magnitude is
v2
.
r
Using the free-body diagram,
X
i
X
i
Fx = N sin θ = m
v2
r
5. h ≈ 30 m
Explanation:
At a height of 10 meters,
y
N
Fc = m ac = m
and
Fy = N cos θ = m g .
Dividing,
(24 m/s)2
v2
=
gr
(9.8 m/s2 )(95.2 m)
= 0.61739
θ = arctan(0.61739) = 31.6908◦ .
tan θ =
005 10.0 points
A ball is thrown upward. At a height of 10
meters above the ground, the ball has a potential energy of 50 Joules (with the potential
energy equal to zero at ground level) and is
moving upward with a kinetic energy of 50
Joules.
What is the maximum height h reached by
the ball? Consider air friction to be negligible.
50 J = m g h = m g (10 m)
mg = 5 N
and the total energy is 100 Joules. v = 0 at
the maximum height, so the kinetic energy is
0. The total energy will not change, so at the
top,
100 J = m g h = (5 N) h
h = 20 m .
006 10.0 points
A time-varying net force acting on a 4.2 kg
particle causes the object to have a displacement given by
x = a + b t + d t2 + e t3 ,
where a = 1.7 m , b = 1.6 m/s , d =
−1.7 m/s2 , and e = 1.2 m/s3 , with x in
meters and t in seconds.
Find the work done on the particle in the
first 3.7 s of motion.
1. 1070.04
2. 161.978
3. 3075.74
4. 298.142
5. 2683.95
6. 2809.64
7. 908.418
8. 240.93
9. 160.154
10. 1610.04
Correct answer: 3075.74 J.
Explanation:
1. h ≈ 50 m
2. h ≈ 10 m
3. h ≈ 20 m correct
4. h ≈ 40 m
3
Let :
ti = 0 s and
tf = 3.7 s .
Since the force is time dependent
Z xf
Z xf
~
m a dx
F · d~x =
W ≡
xi
xi
Version 097 – Exam 2 – hoffmann – (57870)
Z xf
Z xf
dv
dv dx
I
=m
dx = m
dx
dt
dx
dt
4. Zero, since fk · dr = 0
xi
Zxivf
1
1
v dv = m vf2 − m vi2 .
=m
5. 2 π L fk
2
2
vi
π
Therefore work done on the particle is the
+ 1 L fk correct
6. −
2
change in kinetic energy. For this case,
W = ∆K = Kf − Ki =
7. 2 (π + 1) L fk
1
m (vf2 − vi2 ) ,
2
8. 2 L fk
where the velocity is
Explanation:
In the motion from P to Q,
Z
W = fk · dr = −fk L .
dx
= b + 2 d t + 3 e t2 .
dt
vi = 1.6 m/s and
v=
vf = (1.6 m/s) + 2 (−1.7 m/s2 ) (3.7 s)
3
= 38.304 m/s ,
+3 (1.2 m/s ) (3.7 s)
so
2
1
W = Kf − Ki = m (vf2 − vi2 )
2
h
i
1
= (4.2 kg) (38.304 m/s)2 − (1.6 m/s)2
2
= 3075.74 J .
007 10.0 points
On a horizontal surface, an object slides from
point P to point Q along a straight line, then
slides back from Q to P along a semicircle.
P
Q
If P and Q are a distance L apart, how
much work did the force of kinetic friction do
during the entire process? The magnitude of
fk is fk .
1. −(2 π + 1) L fk
2. (2 π − 1) L fk
3. −(π + 1) L fk
4
In the motion from Q back to P , the path
π L fk
L
.
length is π , so W = −
2
2
Thus the total work done by the force of
kinetic friction is
π
−
+ 1 L fk
2
and fk is not a conservative force.
008 10.0 points
A constant force is exerted for a short distance
on a block of mass that is initially at rest on a
horizontal frictionless plane. This force gives
the block a certain final speed v. Suppose we
repeat the experiment but, instead of starting
from rest, the block is already moving with
constant speed 2 v in the direction of the force
at the moment we begin to apply the force.
After we exert the same constant force for
the same distance, the increase in the block’s
speed
√
1. ( 5 + 2) v
2. v
√
3. ( 3 − 1) v
√
4. ( 5 − 2) v correct
5. 0
√
6. ( 3 − 2) v
m
.3
11
7. cannot be determined from the information provided.
√
8. ( 2 − 1) v
5
4m
Version 097 – Exam 2 – hoffmann – (57870)
2.3 kg
Explanation:
40 ◦
1
1
WF = ∆ K = m vf2 − m vi2
2
2
When the block starts from rest, the work
done by the applied force is
vi = 0 ,
vf = v ,
1
WF = m v 2
2
When the block starts from initial velocity v,
vi = 2 v ,
vf = 2 v + ∆v ,
1
WF = m (2 v + ∆v)2 −
2
1
1
m v 2 = m (2 v + ∆v)2 −
2
2√
∆v = ( 5 − 2) v .
α
9.8 m/s2
What is the maximum angle α which the
pendulum will swing to the right after hitting
the peg as shown above?
1. 71.0778
2. 75.6516
3. 59.7175
4. 76.7362
5. 50.3683
6. 68.1834
7. 62.4173
8. 57.5351
9. 77.9371
10. 69.3478
Correct answer: 50.3683◦ .
Explanation:
1
m (2 v)2
2
1
m (2 v)2
2
009 10.0 points
Note: The pendulum bob is released at a
height below the height of the peg.
A pendulum made of a string of length
11.3 m and a spherical bob of mass 2.3 kg
and negligible raius swings in a vertical plane.
The pendulum is released from an angular
position 40 ◦ from vertical as shown in the
figure below. The string hits a peg located
a distance 4 m below the point of suspension
and swings about the peg up to an angle α
on the other side of the peg. Then, the bob
proceeds to oscillate back and forth between
these two angular positions.
The acceleration of gravity is 9.8 m/s2 .
Let : m = 2.3 kg ,
ℓ = 11.3 m ,
d = 4 m,
θ = 40 ◦ , and
g = 9.8 m/s2 .
The radius of the sphere about the peg
r = ℓ − d = 7.3 m .
(1)
Ui = −m g ℓ cos θ .
(2)
Uf = −m g d − m g r cos α .
(3)
m g ℓ cos θ = m g d + m g r cos α
ℓ cos θ − d = r cos α , so
(4)
Relative to the the point of suspension, the
initial potential energy is
The final potential energy at the maximum
angle α which the pendulum will swing to the
right after hitting the peg is
Setting Eqs. 2 and 3 equal, we have
Version 097 – Exam 2 – hoffmann – (57870)
6
ℓ cos θ − d
α = arccos
(5)
011 10.0 points
r
◦
A
block
of
mass
m is pushed a distance D up
(11.3 m) cos 40 − 4 m
= arccos
an inclined plane by a horizontal force F . The
(7.3 m)
plane is inclined at an angle θ with respect to
= 50.3683 ◦ .
the horizontal. The block starts from rest and
the coefficient of kinetic friction is µk .
010 10.0 points
A spring-loaded gun can fire a projectile to a
height h if it is fired straight up.
If the same gun is pointed at an angle of
45◦ from the vertical, what maximum height
h45◦ can now be reached by the projectile?
h
4
h
= correct
2
h
=√
2
h
= √
2 2
1. h45◦ =
2. h45◦
3. h45◦
4. h45◦
5. h45◦ = h
Explanation:
From conservation of energy, when fired
straight up,
1
m v02 = m g h
2
v2
h= 0 .
2g
When fired at an angle of 45◦ ,
v0x = v0y = v0 sin 45 =
◦
√
2
v0
2
and
1
1
m v02 = m
2
2
√
2
v0
2
!2
+ m g h45◦
since the projectile still has horizontal velocity
at the maximum height. Thus
K=U
1
m v02 = m g h45◦
4
v2
1
h45◦ = 0 = h .
4g
2
D
F
m
µk
θ
If N is the normal force, the final speed of
the block is given by
r
2
(F cos θ − m g sin θ) D
1. v =
m
r
2
2. v =
(F cos θ + m g sin θ − µk N ) D
m
r
2
(F cos θ − m g sin θ − µk N ) D
3. v =
m
correct
r
2
(F sin θ + µk N ) D
4. v =
m
r
2
(F sin θ − µk N ) D
5. v =
m
r
2
(F cos θ − m g sin θ + µk N ) D
6. v =
m
r
2
(F cos θ − µk N ) D
7. v =
m
r
2
8. v =
(F cos θ + m g sin θ) D
m
Explanation:
The force of friction has a magnitude
Ff riction = µk N . Since it is in the direction opposite to the motion, we get
Wf riction = −Ff riction D = −µk N D.
The normal force makes an angle of 90◦
with the displacement, so the work done by it
is zero.
Version 097 – Exam 2 – hoffmann – (57870)
The work done by gravity is
Wgrav = m g D cos(90◦ + θ)
= −m g D sin θ .
since θ = 0◦ ⇒ cos θ = 1.
Given:
m = 2.7 × 103 kg
F1 = 1154 N
F2 = 943 N
vf = 2.7 m/s
The work done by the force F is
WF = F D cos θ .
From the work-energy theorem we know that
Wnet = ∆K ,
WF + Wgrav + Wf riction =
1
m vf2 .
2
Thus
vf =
r
2
(F cos θ − m g sin θ − µk N ) D .
m
012 10.0 points
A 2.7 × 10 kg car accelerates from rest under
the action of two forces. One is a forward
force of 1154 N provided by traction between
the wheels and the road. The other is a 943 N
resistive force due to various frictional forces.
How far must the car travel for its speed to
reach 2.7 m/s?
1. 33.6
2. 71.4261
3. 80.9261
4. 72.2179
5. 35.8278
6. 50.5443
7. 57.838
8. 41.0063
9. 24.91
10. 46.6422
7
Solution:
1
(F1 − F2 ) d = mvf2
2
mvf2
d=
2(F1 − F2 )
(2700 kg)(2.7 m/s)2
d=
2(1154 N − 943 N)
= 46.6422 m
3
Correct answer: 46.6422 m.
Explanation:
Basic Concepts:
1
Wnet = ∆K = Kf − Ki = mvf2
2
1
since vi = 0 m/s ⇒ Ki = mvi2 = 0 J.
2
Fnet = F1 − F2
Wnet = Fnet d cos θ = Fnet d
013 10.0 points
A block starts at rest and slides down a frictionless track except for a small rough area on
a horizontal section of the track (as shown in
the figure below).
It leaves the track horizontally, flies through
the air, and subsequently strikes the ground.
The acceleration of gravity is 9.81 m/s2 .
9.81 m/s2
554 g
v
2.3 m
h
b b b b
b b
1.1 m
µ=0.3
b
b
b
b
b
b
3.92 m
At what height h above the ground is the
block released?
1. 4.20337
2. 4.30026
3. 4.10219
4. 4.39526
5. 4.60001
6. 5.20232
7. 4.90073
8. 3.79761
9. 4.99735
10. 3.9005
Version 097 – Exam 2 – hoffmann – (57870)
vx2 = 2 g h1 − 2 µ g ℓ
Correct answer: 4.30026 m.
Explanation:
Let : x = 3.92 m ,
g = 9.81 m/s2 ,
m = 554 g ,
µ = 0.3 ,
ℓ = 1.1 m ,
h2 = −2.3 m ,
h = h1 − h2 , and
vx = v .
h2
ℓ
b
b
b
b
b
Conservation of Me-
Ui = Uf + K f + W .
(1)
since vi = 0 m/s.
1
m v2
2
(2)
Ug = m g h
(3)
W = µmgℓ.
(4)
K=
(6)
x = vx t .
(5)
Choosing the point where the block leaves the
track as the origin of the coordinate system,
∆x = vx ∆t
(5)
1
h2 = − g ∆t2
2
(6)
since axi = 0 m/s2 and vyi = 0 m/s.
Solution: From energy conservation Eqs. 1,
2, 3, and 4, we have
1
m vx2 = m g (h − h2 ) − µ m g ℓ
2
x
from
vx
so
(8)
Using Eq. 6 and substituting vx2 from Eq. 8,
we have
g x2
+ µℓ
2 h2 2 g
x2
+ µℓ
=−
4 h2
(3.92 m)2
=−
+ (0.3) (1.1 m)
4 (−2.3 m)
= 2.00026 m , and
h = h1 − h2
= (2.00026 m) − (−2.3 m)
h1 = −
x
Basic Concepts:
chanical Energy
1
h2 = − g t2
2
2
1
x
h2 = − g
,
2
vx
g x2
.
vx2 = −
2 h2
g
h1
h
b
(7)
Eq. 5, we have
v
b b b b
b b
vx2
h1 =
+µℓ
2g
Using Eq. 6 and substituting t =
m
µ
8
(9)
= 4.30026 m .
014 10.0 points
A penny of mass 3.1 g rests on a small 21 g
block supported by a spinning disk of radius
24 cm. The coefficients of friction between
block and disk are 0.782 (static) and 0.64
(kinetic) while those for the penny and block
are 0.629 (static) and 0.45 (kinetic).
Disk
Penny
r
Block
Version 097 – Exam 2 – hoffmann – (57870)
What is the maximum speed of the disk
without the block and penny sliding? The
acceleration of gravity is 9.8 m/s2 .
1. 73.2251
2. 48.3954
3. 58.4543
4. 47.7979
5. 61.7117
6. 50.9516
7. 65.4945
8. 54.6429
9. 81.5057
10. 62.4001
Correct answer: 48.3954 rpm.
Explanation:
Let :
µ = 0.629 ,
r = 24 cm , and
g = 9.8 m/s2 .
The frictional force provides the centripetal
force
µmg =
m v2
,
r
9
ω
8.8 kg
7.2 m
2.1 kg
Use conservation of energy to determine the
final speed of the first mass after it has fallen
(starting from rest) 7.2 m .The acceleration of
gravity is 9.8 m/s2 .
1. 9.31362
2. 3.86876
3. 6.22231
4. 4.96989
5. 7.2051
6. 8.16905
7. 5.50119
8. 5.31263
9. 8.04009
10. 5.58407
Correct answer: 9.31362 m/s.
Explanation:
so the speed is
a
T
2.1 kg
m2 g
T
8.8 kg
m1 g
015 10.0 points
A 8.8 kg mass is attached to a light cord that
passes over a massless, frictionless pulley. The
other end of the cord is attached to a 2.1 kg
mass.
and
Consider the free body diagrams
a
√
v = µgr = rω
r
µg
ω=
r
r
0.629 (9.8 m/s2 ) 60 s 1 rev
=
0.24 m
1 min 2π
= 48.3954 rpm .
Let : m1 = 8.8 kg ,
m2 = 2.1 kg ,
ℓ = 7.2 m .
Let the figure represent the initial configuration of the pulley system (before m1 falls
down).
Version 097 – Exam 2 – hoffmann – (57870)
From conservation of energy
K i + Ui = K f + Uf
0 + m1 g ℓ = m2 g ℓ +
(m1 − m2 ) g ℓ =
1
1
m1 v 2 + m2 v 2
2
2
1
(m1 + m2 ) v 2
2
Therefore
v=
s
=
s
(m1 − m2 )
2gℓ
(m1 + m2 )
8.8 kg − 2.1 kg
8.8 kg + 2.1 kg
q
× 2 (9.8 m/s2 )(7.2 m)
= 9.31362 m/s .
keywords:
10