Version 048 – Test 2 – swinney – (58385)
This print-out should have 20 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
In each diagram below, a block is acted on by
a force with a magnitude F in the direction
shown. If each block undergoes the same
displacement to the right and only to the
right, which of the following is the correct
order of the amount of work done by force F ,
from most positive to most negative?
~
F
1
9. c, a, b, d
10. d, c, a, b
Explanation:
Wa
Wb
Wc
Wd
=0
= −F ∆x
= F ∆x
= −F cos θ ∆x
Here, F is the magnitude of the force, ∆x
is the magnitude of the displacement, and θ is
the angle of force in (d) relative to horizontal
direction. So, Wc > Wa > Wd > Wb .
002 10.0 points
A block of density ρ1 and volume V1 is submerged in a liquid of density ρL . A second
block of density ρ2 and volume V2 is placed
on top of the first block. The two blocks are
floating in the manner shown in the figure
below.
(a)
~
F
(b)
~
F
(c)
V2
~
F
V1
(d)
1. a, d, b, c
2. d, a, b, c
3. a, b, c, d
4. b, d, a, c
5. b, c, d, a
6. a, c, b, d
7. c, b, d, a
8. c, a, d, b correct
liquid
Find V2 such that the two blocks are just
submerged, as shown in figure above.
ρL − ρ1
1. V2 =
V1
ρL − ρ2
ρL − ρ2
V1
2. V2 =
ρL − ρ1
ρL + ρ1
V1
3. V2 =
ρL + ρ2
ρL
4. V2 =
V1
ρ2 + ρ1
ρ2 + ρ1
V1
5. V2 =
ρL
ρL + ρ1
V1
6. V2 =
ρ2
Version 048 – Test 2 – swinney – (58385)
ρL − ρ1
=
ρ − ρL
2
ρL − ρ2
=
ρ − ρL
1
ρ2
=
ρ − ρ1
L
ρL + ρ2
=
ρL + ρ1
2
defined as
P
i mi xi
xCM = P
i mi
8. V2
V1
m1 x1 + m2 x2
=
m0 + m1 + m2
V1
9. V2
(8.2303 kg) (7.04 cm) + (2.57623 kg) (20.9792 cm)
=
(4.87 kg) + (8.2303 kg) + (2.57623 kg)
V1
10. V2
= 7.14371 cm
P
Explanation: X
i mi yi
In equilibrium
Fy = 0, so the buoyant yCM = P
i mi
force is
m1 y1 + m2 y2
=
m0 + m1 + m2
(8.2303 kg) (14.08 cm) + (2.57623 kg) (0 cm)
FB = ρL VT g
=
(4.87 kg) + (8.2303 kg) + (2.57623 kg)
m1 g + m2 g = ρL (V1 + V2 ) g
= 7.39211 cm
ρ1 V1 + ρ2 V2 = ρL V1 + ρL V2
ρL − ρ1
and
V2 =
V1 .
q
ρ2 − ρL
2
rCM = x2CM + yCM
q
003 10.0 points
= (7.14371 cm)2 + (7.39211 cm)2
Three point mass particles are located in a
= 10.2799 cm
plane:
4.87 kg located at the origin,
8.2303 kg at [(7.04 cm), (14.08 cm)],
004 10.0 points
and
A ball of mass m is swinging in a horizontal
2.57623 kg at [(20.9792 cm), (0 cm)].
circle of radius r at the end of a string of
How far is the center of mass of the three
length ℓ = 4 r.
particles from the origin?
1. 9.31585
2. 5.98445
3. 7.76919
4. 10.4535
θ
5. 8.41693
6. 9.19941
ℓ = 4r
7. 10.2799
8. 7.63386
9. 8.76571
10. 4.81256
r
Correct answer: 10.2799 cm.
v
Explanation:
If the ball swings at constant speed, what is
Basic Concepts: Center of mass
its
speed in terms of the magnitude T of the
P
m
~
r
tension in the string, as well as r and m?
i i
~rCM = Pi
.
m
r
i i
Tr
1.
Solution: The center of mass (xCM , yCM ) is
m
7. V2
V1 correct
Version 048 – Test 2 – swinney – (58385)
2. The speed does not depend on T , but
rather on m, g and r.
r
T
3. 4
mr
r
r
T
4.
4 m
r
1 Tr
5.
correct
2
m
r
T
6. 2
mr
r
T
1
7.
2 mr
r
T
8.
mr
Tr
9.
m
Explanation:
From Newton’s 2nd Law,
m v2
r
T
m ar =
= T sin θ = T
=
r
4r
4
Tr
2
mv =
r
r4
Tr
1 Tr
=
.
v=
4m
2
m
3
5. W = 0
6. W =
A2 D 3
− A B D2 + B 2 D correct
3
7. W = 2 A B D4
8. W = (A D − B) D
Explanation:
W =
W =
Z
Z
F · dr ,
so
D
(A x − B)2 dx
0
D
A 2 x3
2
2
=
− A B x + B x 3
0
2
A D3
− A B D2 + B 2 D .
=
3
006 10.0 points
Five ramps lead from the ground to the second
floor of a workshop, as sketched below. All
five ramps have the same height; ramps B,
C, D and E have the same length; ramp A is
longer than the other four. You need to push
a heavy cart up to the second floor and you
may choose any one of the five ramps.
005 10.0 points
A force acts along the x axis and is described
by
F(x) = î (A x − B)2 ,
where A and B are positive constants.
How much work does this force do on a
particle moving from x = 0 to x = D?
A
B
C
D
A B D2
1. W =
2
E
2. W = D (A D − B)
2
2
3. W = (A D − B) − B A D
2
A
1
A
+
+
DB 2 DB
2
A D2
4. W =
−BD
2
Assuming no frictional forces on the cart,
which ramp would require you to do the least
work?
1. Ramp C.
2. Ramp A.
Version 048 – Test 2 – swinney – (58385)
3. Ramp B.
4. Same work for ramps B, C, D or E; more
work for ramp A.
4
2.
3.
5. Same work for all five ramps. correct
6. Same work for the straight ramps A and
B; less work for ramps C, D, and E.
4.
7. Unable to determine without knowing the
exact profiles of ramps C, D or E.
8. Ramp E.
5.
9. Ramp D.
Explanation:
Let h be the height of the ramps. Since
there is no friction, we can use the workenergy theorem.
6.
7.
Wtot = Wperson + Wgravity = ∆K
Hence for all the ramps,
Wperson = −Wgravity + ∆K
= mgh + ∆K
In particular, if ∆K = 0 (the cart starts from
rest and ends at rest), Wperson = mgh, for all
the ramps.
007 10.0 points
A car is traveling very slowly around a banked
curve.
What is the free body diagram that describes the forces acting on the car?
1.
8.
correct
Explanation:
Since the car is moving very slowly, the
friction force is opposing its tendency to slide
down the incline. Its direction is then parallel to and up the incline. The gravity force
points straight down. The normal force is
perpendicular to the surface.
008 10.0 points
A small piece of wood is dropped from a
height 3 m above water. The wood has density 690 kg/m3 while the water’s density is
1000 kg/m3 .
How deep would the wooden piece sink into
the water before floating back to the surface
if there were no viscosity or any other kind of
water resistance to its motion?
1. 4.5973
2. 10.5276
Version 048 – Test 2 – swinney – (58385)
3. 3.0
4. 5.65714
5. 9.27647
6. 3.51481
7. 6.67742
8. 20.475
9. 8.72857
10. 2.03333
5
U(x)
B
A
Correct answer: 6.67742 m.
−x0 0
x0 2x0 3x0 4x0 5x0
x
Explanation:
What is the force F on the particle at x =
3 x0 ?
3
Let : ρwood = 690 kg/m ,
ρwater = 1000 kg/m3 ,
h = 3 m.
and
The wooden piece starts its downward journey from rest and when its reaches the maximal depth d, its speed again drops to zero. By
the work-energy theorem, the net work of all
the forces acting on the wooden piece must be
2. F
3. F
4. F
Wnet = ∆Ekin = 0 − 0 = 0 .
In the absence of water resistance of any kind,
there are only two forces acting on the piece,
its weight M g acting all the time, and the
buoyant force Fb that acts only when the piece
is underwater, so
Wnet = m g (h + d) − B d = 0 ,
mgh
mgh
=
B − mg
V ρwater g − m g
mgh
=
ρwater
− mg
mg
ρwood
h ρwood
=
ρwater − ρwood
(3 m) (690 kg/m3 )
=
1000 kg/m3 − 690 kg/m3
d=
= 6.67742 m .
009 10.0 points
The graph is a potential energy diagram for
a particle in some system as a function of
position x.
B
5 x0
B−A
=
4 x0
B−A
=−
4 x0
B
=
correct
4 x0
A
=−
4 x0
B
=−
4 x0
1. F =
5. F
6. F
7. F = 0
A
5 x0
B
9. F = −
5 x0
A
10. F =
4 x0
8. F =
Explanation:
F =−
0−B
B
dU
=−
=
.
dr
5 x0 − x0
4 x0
010 10.0 points
Planet A and planet B of equal mass m orbit
the same star of mass M in circular trajectories of radii rA = R and rB = 2R respectively.
Calculate the ratio of the kinetic energy of A
to the kinetic energy of B.
Version 048 – Test 2 – swinney – (58385)
1. 16
3. 8
4. 1/16
5. 2 correct
6. 1/8
7. 1/4
8. 4
Explanation:
Since A has a circular orbit,
2
m vA
GmM
=
,
R
R2
which implies
r
gR
correct
2
p
5. v = 2 2 g R
p
6. v = g R
r
gR
7. v = 2
2
p
1
2gR
8. v =
2
r
1 gR
9. v =
2
2
Explanation:
The net applied force in the radial direction
is
4. v =
2. 1/2
Frad = m g − N,
where N is the normal force exerted on the
driver by the seat. It is given that
1
GmM
2
KEA = m vA
=
.
2
2R
N=
2
m vB
GM m
=
,
2R
4 R2
which implies
Frad =
1
m g.
2
Now, by the Momentum Principle
1
GM m
2
KEB = m vB
=
.
2
4R
KEA
= 2.
KEB
011 10.0 points
What must the speed v of car be as it goes
over the top of a hill for the driver to feel half
as heavy as normal? Let R be the radius of
the kissing circle at the top of the hill.
p
1. v = 2 g R
1p
2. v =
gR
2
p
3. v = 2 g R
1
m g,
2
so
Similarly, for B,
Therefore,
6
Frad =
m v2
.
R
Setting
1
m v2
mg =
.
2
R
finally gives
v=
r
gR
.
2
012 10.0 points
A rock on a string is whirled fast enough to
move in a vertical circle as shown. Gravity is
acting downward.
Version 048 – Test 2 – swinney – (58385)
7
Explanation:
Let : θ = 315◦ .
The mass m of the rock is not required and
the length R of the string is not required.
The free body diagram is
R
center
rock
Fnet
T
What is the direction of the rock’s acceleration at the position shown?
1.
mg
The tension along the string must point
toward the center of the circle and the gravitational force is down, therefore we have
X
Fr : T − m g cos θ = m ar
X
Fθ :
m g sin θ = m aθ .
2.
3. None of these
4.
Thus, the free body diagram gives us the
centripetal acceleration ar and the tangential
acceleration aθ
v2
T
− g cos θ =
and
(1)
ar =
m
R
aθ = g sin θ .
(2)
5.
The tension in the string T is
v2
T =m
+ m g cos θ > 0 .
R
6.
v2
should always be larger than g cos θ since
R
the rock has just enough velocity to make a
full vertical circle.
The vector sum of the radial acceleration ar
and tangential acceleration aθ is shown below.
7.
8.
a
correct
ar
9.
10.
aθ
Version 048 – Test 2 – swinney – (58385)
At the top of the circle θ = 180◦ and T ≈ 0,
8
The magnitude is
so
v2
≈ g,
R
≈ gR.
ar =
2
vtop
|α| = 9.10081◦ .
or
(3)
At the bottom of the circle θ = 0◦ and T is
maximum, so
1
1
2
2
m vbottom
+ mgh
= m vtop
2
2
1
= m g R + m g (2 R) ,
2
p
vbottom = 5 g R .
so
013 10.0 points
Two blocks of the same size, shape, and mass
m are stacked one atop the other. It is desired
to push the blocks at constant acceleration
across a level tabletop where there is kinetic
friction.
1
(4)
F
µs
Using Eq. 3, h′ = R (1 + cos θ) , and the
conservation of energy, at an angle θ = 315◦ ,
we have
1
1
2
m v 2 = m vtop
+ m g h′
2
2
1
= m g R + m g (1 + cos θ) R ,
2
v2
= g + 2 g (1 + cos θ) , so
R
ar = g (3 + 2 cos θ)
(1)
2
◦
= 9.8 m/s (3 + cos 315 )
= 43.2593 m/s2 , and
aθ = g sin θ
= 9.8 m/s2 sin 315◦
(2)
= −6.92965 m/s2 .
Using the acceleration vector diagram and
Eqs. 1 and 2, we have
q
a = a2r + a2θ
q
= (43.2593 m/s2 )2 + (−6.92965 m/s2 )2
= 43.8108 m/s2 .
Using the acceleration vector diagram and
Eqs. 1 and 2, we have
aθ
α = arctan
ar
−6.92965 m/s2
= arctan
43.2593 m/s2
◦
= −9.10081
m
2
m
µk
If the blocks are to move together, what
minimum coefficient of static friction must
exist between the two blocks, if a constant
horizontal force F is applied to the upper
block? The answer should depend only on F ,
m, g and µk (for the kinetic friction between
the lower block and table).
1. µs = 2µk
F
− µk
mg
2F
=
+ µk
mg
F
=
2mg
F
=
+ µk correct
2mg
F
=
+ 2µk
mg
F
+ µk
=
mg
F
=2
+ µk .
mg
2. µs =
3. µs
4. µs
5. µs
6. µs
7. µs
8. µs
9. µs = 0, since F accelerates both blocks.
Version 048 – Test 2 – swinney – (58385)
10. µs = µk
Explanation:
n1
1
F
m
n1
2
a
fs
fs
m
a
fk
mg
n mg
For the upper block, apply Newton’s second
law horizontally:
m a = F − fs
and vertically:
0 = n1 − m g
where
fsmax = µs n1 = µs m g .
For the lower block, apply Newton’s second
law horizontally:
m a = fs − fk
and vertically:
9
014 10.0 points
A crate with a mass of 85 kg glides through a
space station with a speed of 3.4 m/s. Sandra
Bullock speeds it up by pushing on it from
behind with a force of 185 N, continually
pushing with this force through a distance
of 5 m. Then Sandra moves around to the
front of the crate and slows the crate down
by pushing backward with a force of 175 N,
backing up through a distance of 3 m. After
these two maneuvers, what is the speed of the
crate?
1. 3.45581
2. 4.34927
3. 4.24971
4. 4.97092
5. 4.57949
6. 4.68963
7. 4.12647
8. 4.02295
9. 5.45985
10. 4.88383
Correct answer: 4.57949 m/s.
Explanation:
First we need to find the total work done
on the system. We assume the forces on the
crate are parallel to its motion:
0 = n − n1 − m g = n − (2 m) g
where
fk = µk n = µk (2 m) g .
Adding the horizontal equations,
2 m a = F − f k = F − 2 µk g
F
a=
− µk g .
2m
But
fs = F − m a =
F
+ µk m g ,
2
so for the maximum force of static friction,
F
+ µk m g
2
F
µs =
+ µk .
2mg
µs m g =
~ 1 ∆~r1 + F
~ 2 ∆~r2
W =F
= F1,x ∆x1 + F2,x ∆x2
= (185 N)(5 m) + (−175 N)(3 m)
= 400 J .
Note that F2,x is negative because at this
point Sandra is pushing opposite to the motion. Now that we know the work, we can find
the final kinetic energy:
KEf = KEi + W
1
= m v2 + W
2
1
= (85 kg)(3.4 m/s)2 + 400 J
2
= 891.3 J .
Now we can find the final speed:
Version 048 – Test 2 – swinney – (58385)
KE =
⇒ vf =
=
10
1
m v2
2
s
2(KEf )
m
s
r
2(891.3 J)
85 kg
θ
O
= 4.57949 m/s .
015 10.0 points
~ = Fx ı̂+Fy ̂ moves
A two-dimensional force F
a particle to a new location that is ~r = rx ı̂ +
ry ̂ from where it was originally located. If
Fx = 9 N, Fy = −5 N, rx = 3 m, and ry =
3 m, what is the work done on the particle by
the force?
1. -5.0
2. 12.0
3. 8.0
4. 44.0
5. -4.0
6. 6.0
7. 5.0
8. 4.0
9. 24.0
10. 0.0
The magnitude of a, the total acceleration
is
1. |~a| =
rect
2. |~a| =
3. |~a| =
4. |~a| =
Correct answer: 12 J.
Explanation:
The work is given by
5. |~a| =
6. |~a| =
~ · ∆~r = Fx rx + Fy ry
W =F
= (9 N) (3 m) + (−5 N) (3 m)
7. |~a| =
= 12 J .
8. |~a| =
016 10.0 points
A small sphere of mass m is connected to the
end of a cord of length r. It rotates in a vertical circle about a fixed point O while being
acted on downward by gravity which has the
acceleration constant g. Let the magnitude of
the tension force exerted by the cord on the
sphere be denoted by T .
9. |~a| =
10. |~a| =
s
s
s
s
s
s
s
s
s
s
T
m
T
m
T
m
T
m
T
m
T
m
T
m
T
m
T
m
T
m
Explanation:
2
2
2
2
2
2
2
2
2
2
+ 2 g sin θ
T
+ g 2 . corm
cos2 θ + g 2 sin2 θ .
+ 2 g cos θ
T
+ g2 .
m
− g2 .
+ g 2 sin2 θ .
− 2 g sin θ
T
+ g2 .
m
+ g 2 cos2 θ .
− 2 g cos θ
T
+ g2 .
m
sin2 θ + g 2 cos2 θ .
+ g2 .
Version 048 – Test 2 – swinney – (58385)
ar
θ
at
11
A block of mass m is pushed a distance D up
an inclined plane by a horizontal force F . The
plane is inclined at an angle θ with respect to
the horizontal. The block starts from rest and
has velocity is vf when the block is at D.
D
O
F
m
µk
θ
The coefficient of kinetic friction µk is
The centripetal force is
1. µk =
m v2
Fc =
.
r
This centripetal force is provided by the tension force and the radial component of the
weight. In this case, they are in opposite
directions so,
Fc =
m v2
= T + m g sin θ .
r
Then, we see that the radial acceleration is
T
+ g sin θ .
ar =
m
The tangential acceleration is caused by the
tangential component of the weight, so
at = g cos θ .
Thus the total acceleration is
q
a = a2r + a2t
" 2
T
T
=
+ 2 g sin θ
m
m
12
2
2
2
+ g (sin θ + cos θ)
=
s
T
m
017
2
+ 2 g sin θ
T
+ g2 .
m
10.0 points
F D cos θ − m g D cos θ − 1/2m vf2
m g D sin θ + F D cos θ
F D cos θ − m g sin θ
m g cos θ + F sin θ
F D cos θ + m g sin θ
3. µk =
m g cos θ − F sin θ
F D cos θ − m g D sin θ − 1/2m vf2
4. µk =
m g D cos θ
F D cos θ − m g sin θ
5. µk =
m g cos θ − F sin θ
F D sin θ − m g cos θ − 1/2m vf2
6. µk =
m g sin θ
F cos θ − m g sin θ − 1/2m vf2
7. µk =
m g cos θ + F sin θ
F D cos θ − m g D sin θ − 1/2m vf2
8. µk =
m g D cos θ + F D sin θ
correct
2. µk =
9. µk =
10. µk =
F D sin θ − m g sin θ − 1/2m vf2
m g cos θ
m g D sin θ + 1/2m vf2 − F D cos θ
m g D cos θ
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.
= −µk (m g cos θ + F sin θ) D
Version 048 – Test 2 – swinney – (58385)
The normal force makes an angle of 90◦
with the displacement, so the work done by it
is zero.
The work done by gravity is
Wgrav = −m g D sin θ .
The work done by the force F is
5. vmin =
s
g R (cos θ + µ sin θ)
sin θ + µ cos θ
6. vmin =
s
g R (sin θ − µ cos θ)
correct
cos θ + µ sin θ
7. vmin =
s
g R (sin θ + µ cos θ)
cos θ − µ sin θ
8. vmin =
s
g R (cos θ − µ sin θ)
sin θ + µ cos θ
9. vmin =
s
g R (sin θ − µ sin θ)
cos θ + µ cos θ
10. vmin =
s
g R (sin θ + µ cos θ)
cos θ + µ sin θ
WF = F D cos θ .
From the Energy Principle we know that
Wnet = ∆K ,
WF + Wgrav + Wf riction
1
= m vf2 .
2
Thus
µk =
F D cos θ − m g D sin θ − 1/2m vf2
m g D cos θ + F D sin θ
018 10.0 points
A car rounds a curve. The radius of curvature
of the road is R, the banking angle with respect to the horizontal is θ and the coefficient
of static friction is µ.
12
Explanation:
Since we want to calculate the minimum
speed in order for the car to not slip, we need
to consider the case where the frictional force
is Ff = µ N and is directed up the incline.
There is no acceleration in the vertical direction and thus
N cos θ+Ff sin θ = N cos θ+µ N sin θ = m g .
Solving for the normal force, we get
M
N =
µ
θ
mg
.
cos θ + µ sin θ
In the horizontal direction we have
N sin θ − Ff cos θ = N (sin θ − µ cos θ)
What is the minimum speed that the car
can have without slipping assuming that if
the car were at rest, it would slide down the
plane?
s
g R (cos θ + µ cos θ)
1. vmin =
sin θ − µ sin θ
s
g R (sin θ − µ cos θ)
2. vmin =
cos θ − µ sin θ
s
g R (cos θ + µ sin θ)
3. vmin =
sin θ − µ cos θ
s
g R (cos θ − µ sin θ)
4. vmin =
sin θ − µ cos θ
=m
2
vmin
,
R
and then using the expression we obtained for
N,
g R(sin θ − µ cos θ)
2
vmin
=
cos θ + µ sin θ
s
g R (sin θ − µ cos θ)
.
⇒ vmin =
cos θ + µ sin θ
019
10.0 points
A block slides up a frictionless incline, slows
to a stop and slides back down.
Version 048 – Test 2 – swinney – (58385)
F
G
o
cti
e
r
di
no
f
ti
mo
Car B has the same mass as Car A but is
going twice as fast.
If the same constant force brings both cars
to a stop, how far will car B travel in coming
to rest as compared to the distance car A
traveled?
D
E
C
B
on
13
1. Four times as far correct
A
2. A quarter as far
Which of the following represents the direction of the ball’s acceleration at the times
indicated?
A
1.
B
C
D
E
F
G
3. Three times as far
4. None of these
5. Twice as far
6. Eight times as far
2.
A
B
C
D
E
F
G
7. Insufficient information is given.
8. Half as far
A
B
C
A
B
C
5.
A
B
C
6.
A
B
C
7.
A
B
C
3.
4.
correct
D
E
F
G
E
F
G
D
E
F
G
D
E
F
G
E
F
G
D
D
Explanation:
∆ ~v
~a =
, so ~a is in the same direction as
∆t
∆ ~v. By subtracting the velocity vectors, one
can see that the direction of ∆ ~v is exactly
opposite the direction of ~v . Therefore, the
direction of ~a is down the incline at all points.
020
10.0 points
9. The same distance
Explanation:
The work done by friction in stopping the
cars is W = f d = ∆K . Since the force of
friction is proportional to the mass of the car,
as is K, then d only depends on v 2 . Thus, if
v is doubled, the stopping distance increases
by 4 times. (Standard driver’s license exam
question!)