Version PREVIEW – Practice 7 – carroll – (11108)
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Atwood Machine 05
001 10.0 points
A 6.3 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 3.4 kg
mass.
The acceleration of gravity is 9.8 m/s2 .
1
From conservation of energy
Ki + Ui = Kf + 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
(m1 − m2 )
2gℓ
(m1 + m2 )
6.3 kg − 3.4 kg
=
6.3 kg + 3.4 kg
2
× 2 (9.8 m/s )(5.6 m)
ω
= 5.72842 m/s .
1/2
6.3 kg
5.6 m
3.4 kg
Use conservation of energy to determine the
final speed of the first mass after it has fallen
(starting from rest) 5.6 m .
Correct answer: 5.72842 m/s.
Explanation:
Let : m1 = 6.3 kg ,
m2 = 3.4 kg ,
ℓ = 5.6 m .
and
6.3 kg
T
vT
3.4 kg
a
a
T
Consider the free body diagrams
Frictionless Vertical Circle
002 (part 1 of 3) 10.0 points
A block of mass 0.5 kg is pushed against a horizontal spring of negligible mass, compressing
the spring a distance of ∆x as shown in the figure. The spring constant is 276 N/m. When
released, the block travels along a frictionless,
horizontal surface to point B, the bottom of
a vertical circular track of radius 0.9 m, and
continues to move up the track. The speed
of the block at the bottom of the track is
15 m/s, and the block experiences an average frictional force of 5 N while sliding up the
track.
The acceleration of gravity is 9.8 m/s2 .
T
m2 g
m1 g
R
Let the figure represent the initial configuration of the pulley system (before m1 falls
down).
vB
B
What is ∆x?
Correct answer: 0.638442 m.
∆x
m
k
Version PREVIEW – Practice 7 – carroll – (11108)
Explanation:
From conservation of energy, the initial potential energy of the spring is equal to the
kinetic energy of the block at B. Therefore,
we write
1
1
2
k (∆x)2 = m vB
2
2
s
=
Since
2 ET
− 2 g hT ,
m
2 (42.1128 J)
=
0.5 kg
− 2(9.8 m/s2 ) (1.8 m)
vT 2 =
= 133.171 m2 /s2 ,
2
m vB
k
∆x =
s
(0.5 kg) (15 m/s)2
(276 N/m)
then
vT =
= 0.638442 m .
003 (part 2 of 3) 10.0 points
What is the speed of the block at the top of
the track?
Correct answer: 11.54 m/s.
Explanation:
The change in the total energy of the block
as it moves from B to T is equal to the work
done by the frictional force
∆E = Wf
ET − EB = Wf .
The work done by the frictional force is
Wf = −f π R
= −(5 N) (π) (0.9 m)
= −14.1372 J .
We can find now the speed of the block at T
from
1
m vT 2 = ET − m g hT .
2
133.171 m2 /s2
004 (part 3 of 3) 10.0 points
What is the centripetal acceleration of the
block at the top of the track?
Correct answer: 147.968 m/s2 .
Explanation:
The centripetal acceleration at T is
vT2
R
(11.54 m/s)2
=
0.9 m
ac =
= 147.968 m/s2 .
Bead on a Loop the Loop 01
005 10.0 points
A bead slides without friction around a loopthe-loop. The bead is released from a height
of 26.3975 m from the bottom of the loop-theloop which has a radius 9.999 m.
The acceleration of gravity is 9.8 m/s2 .
Therefore, the total energy at T is
ET = EB + Wf
= 56.25 J + (−14.1372 J)
= 42.1128 J .
q
= 11.54 m/s .
The total energy at B is
1
2
EB = m vB
2
1
= (0.5 kg) (15 m/s)2
2
= 56.25 J .
2
A
9.999 m
26.3975 m
Version PREVIEW – Practice 7 – carroll – (11108)
What is its speed at point A ?
Correct answer: 11.1996 m/s.
3
m
h1
v
b b b b
b b
b
Let :
h2
h
Explanation:
R = 9.999 m and
h = 26.3975 m .
From conservation of energy, we have
Ki + Ui = Kf + Uf
m v2
+ m g (2 R)
2
v 2 = 2 g (h − 2 R) .
b
b
b
b
b
3.6 m
Choose the point where the block leaves the
track as the origin of the coordinate system.
While on the ramp,
0 + mgh =
Therefore
p
v = 2 g (h − 2 R)
r
h
i
= 2 (9.8 m/s2 ) 26.3975 m − 2 (9.999 m)
= 11.1996 m/s .
Block Jump Ramp 01
006 (part 1 of 3) 10.0 points
A block starts at rest and slides down a frictionless track. It leaves the track horizontally,
flies through the air, and subsequently strikes
the ground.
Kb = Ut
1
m vx2 = −m g h1
2
vx2 = −2 g h1
p
vx = −2 g h1
q
= −2 (−9.81 m/s2 ) (1.8 m)
= 5.94273 m/s .
007 (part 2 of 3) 10.0 points
What horizontal distance does the block
travel in the air?
Correct answer: 3.6 m.
Explanation:
441 g
Let :
1.8 m
3.6 m
v
b b b b
b b
b
h2 = −1.8 m .
With the point of launch as the origin,
b
b
b
b
h2 =
b
x
What is the speed of the ball when it leaves
the track? The acceleration of gravity is
9.81 m/s2 .
t=
Let :
x = vx t = vx
g = −9.81 m/s2 ,
m = 441 g , and
h1 = 1.8 m .
2 h2
.
g
Thus
Correct answer: 5.94273 m/s.
Explanation:
1 2
gt
2
s
s
2 h2
g
s
= (5.94273 m/s)
= 3.6 m .
2 (−1.8 m)
−9.81 m/s2
Version PREVIEW – Practice 7 – carroll – (11108)
4
008 (part 3 of 3) 10.0 points
What is the speed of the block when it hits
the ground?
Correct answer: 8.40428 m/s.
ω
Explanation:
4.51 kg
Let :
h = 1.8 m .
4.51 m
2.17 kg
Now choose ground level as the origin.
nergy conservation gives us
Find the speed of each object just as they
pass each other.
Correct answer: 3.93478 m/s.
Kf = Ui
1
m vf2 = −m g h
(7)
2
p
vf = −2 g h
q
= −2 (−9.81 m/s2 ) (1.8 m)
= 8.40428 m/s .
Explanation:
Let : m1 = 4.51 kg ,
m2 = 2.17 kg ,
h = 4.51 m .
−2 (−9.81 m/s2 ) (1.8 m)
= 5.94273 m/s , so
q
(10)
vf = vx2 + vy2
q
= (5.94273 m/s)2 + (5.94273 m/s)2
= 8.40428 m/s .
Serway CP 05 63
009 (part 1 of 3) 10.0 points
Two objects are connected by a light string
passing over a light frictionless pulley as
shown in the figure. The 4.51 kg object is
released from rest at a point 4.51 m above the
floor.
The acceleration of gravity is 9.8 m/s2 .
2.17 kg
4.51 kg
a
T
(9)
a
−2 g h2
m1 g
q
m2 g
=
p
T
Consider the free body diagrams
Alternate Solution:
vy =
and
1
When they meet both are h from the
2
floor, and they have the same speed since
they are connected by the string. Applying
conservation of energy,
(K + Ug )i = (K + Ug )f
1
1
m1 g h = m1 v 2 + m2 v 2
2
2
1
h
+ m1 g
2
1
h
+ m2 g
2
2 m1 g h = (m1 + m2 ) v 2
Version PREVIEW – Practice 7 – carroll – (11108)
+ (m1 + m2 ) g h
m1 g h − m2 g h = (m1 + m2 ) v 2
(m1 − m2 ) g h
v2 =
.
m1 + m2
Thus
s
v=
=
s
5
The 2.17 kg mass becomes a projectile launched straight upward with viy =
5.56463 m/s.
2
vf2y = viy
− 2 g ∆y = 0
at the maximum height, so
(m1 − m2 ) g h
m1 + m2
(4.51 kg − 2.17 kg) (9.8 m/s2 ) (4.51 m)
2.17 kg + 4.51 kg
∆ymax =
2
viy
2g
(5.56463 m/s)2
=
2 (9.8 m/s2 )
= 1.57985 m .
= 3.93478 m/s
010 (part 2 of 3) 10.0 points
b) What is the speed of the each object at the
instant before the 4.51 kg hits the floor?
Correct answer: 5.56463 m/s.
Explanation:
Given : y2f = y1f = h
y1f = 0
Sliding Down a Dome
012 (part 1 of 3) 10.0 points
A small box of mass m is at the top of a
spherical dome with radius R. Starting from
rest after a slight push, the box slides down
along the frictionless spherical surface (see
figure). Ignore the initial kinetic energy (obtained from the slight push).
m
v
θ
R
Conservation of mechanical energy gives
1
(m1 + m2 ) vf2 + m2 g h
2
2 (m1 − m2 ) gh
vf2 =
m + m2
s 1
2 (m1 − m2 ) g h
vf =
m1 + m2
s
2 (4.51 kg − 2.17 kg)
=
2.17 kg + 4.51 kg
q
× (9.8 m/s2 ) (4.51 m)
m1 g h =
= 5.56463 m/s .
011 (part 3 of 3) 10.0 points
c) How much higher does the 2.17 kg object
travel after the 4.51 kg object hits the floor?
Correct answer: 1.57985 m.
Explanation:
Apply the principle of conservation of energy to select the correct equation that relates
the speed v of the box to the polar angle θ and
the radius R while the box is sliding down on
the surface.
p
1. v = 2 g R (1 − cos θ) correct
p
2. v = g R
p
3. v = 2 g R sin θ
p
4. v = 2 g R
p
5. v = g R cot θ
p
6. v = 2 g R (1 − sin θ)
p
7. v = g R sin θ
Version PREVIEW – Practice 7 – carroll – (11108)
8. v =
9. v =
10. v =
r
p
p
g R (1 + tan θ)
2
7. v > g R cot θ
2 g R cos θ
8.
g R tan θ
9. v < g R sin θ
Explanation:
Basic Concepts: Potential Energy, Kinetic Energy, Centripetal Force
m
R
θ
θ
v
mg
Since the surface is frictionless, the mechanical energy of the box is conserved, so the
initial mechanical energy is the same as the
final mechanical energy
1
1
m g R + m vi2 = m g R cos θ + m v 2 .
2
2
It starts from rest (the push is “slight”) so
vi = 0, and we find
1
m g R (1 − cos θ) = m v 2
2
p
v = 2 g R (1 − cos θ) .
013 (part 2 of 3) 10.0 points
Select the inequality relation which corresponds to the condition that the small box
will stay on the surface.
1. v 2 R < g (1 + cos θ)
v2
> g sin θ
R
v2
3.
< g cos θ correct
R
2.
v2
< g sin θ
R
v 2 sin θ
> g (1 + cos θ)
2
Explanation:
10.
2
The radial acceleration of the box is vR
while it is on the surface. So the net radial
m v2
. The normal force the
force on the box is
R
surface exerts on the box can only be outward
so gravity must supply the total force inward,
with any extra counteracted by the normal
force, which must be positive. Thus
m v2
= m g cos θ − N ≤ m g cos θ .
R
Therefore,
v2
≤ g cos θ .
R
014 (part 3 of 3) 10.0 points
Find the critical angle at which the box leaves
the surface of the dome.
1. θ = 36.9◦
2. θ = 41.7◦
3. θ = 48.2◦ correct
4. θ = 49.7◦
5. θ = 46.5◦
6. θ = 39.3◦
7. θ = 51.3◦
4. 2 v 2 > g R (1 + cot θ)
8. θ = 34.1◦
v2
< g R (1 − sin θ)
2
v2
6.
< 2 g tan θ
R
9. θ = 44.1◦
5.
6
10. θ = 38.6◦
Explanation:
Version PREVIEW – Practice 7 – carroll – (11108)
The box leaves the surface when the normal
force is zero. At this instant the component
of the gravity force along the normal to the
surface will be equal to the mass times the
v2
centripetal acceleration
. From the diaR
gram, we see that
m v2
= m g cos θ .
R
(1)
At a height of 10 meters,
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 .
From Part 1:
1
m v 2 = m g R (1 − cos θ)
2
m v2
= 2 m g (1 − cos θ) .
R
(2)
Putting Eqs. (1) and (2) together, we have
m g cos θ = 2 m g (1 − cos θ)
3 m g cos θ = 2 m g
cos θ =
2
3
θ = 48.2◦ .
AP M 1993 MC 6
015 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.
1. h ≈ 30 m
2. h ≈ 20 m correct
3. h ≈ 50 m
4. h ≈ 10 m
5. h ≈ 40 m
Explanation:
7