Version 014 – TEST02 – TSOI – (58160)
1
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printout
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001
10.0 points
An ore car of mass 37000 kg starts from rest
and rolls downhill on tracks from a mine. At
the end of the tracks, 28 m lower vertically, is
a horizontally situated spring with constant
3
.
3
×
10
5
N
/
m.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Ignore friction.
How much is the spring compressed in stop
ping the ore car?
1. 9.86343
2. 6.7462
3. 8.05025
4. 5.34743
5. 5.55889
6. 7.94607
7. 5.00213
8. 5.0898
9. 7.84424
10. 5.63231
Correct answer: 7
.
84424 m.
Explanation:
Energy is conserved, so the change of po
tential energy from when the car is at rest to
when it just hits the spring is
m g h
=
1
2
m v
2
.
The kinetic energy is then converted to po
tential energy in the spring as the cart comes
to rest. When the spring is fully compressed
by an amount
d
, all of the kinetic energy has
been converted to potential energy so
1
2
m v
2
=
1
2
k d
2
.
Thus,
1
2
k d
2
=
m g h ,
and solving for
d
we have
d
=
radicalbigg
2
m g h
k
=
radicalBigg
2 (37000 kg) (9
.
8 m
/
s
2
) (28 m)
(3
.
3
×
10
5
N
/
m)
=
7
.
84424 m
.
002
10.0 points
According to some nineteenthcentury geo
logical theories (now largely discredited), the
Earth has been shrinking as it gradually cools.
If so, how would
g
have changed over geo
logical time?
1.
It would not change;
the mass of the
Earth remained the same.
2.
It would decrease; the Earth’s radius is
decreasing.
3.
It would increase;
g
is inversely propor
tional to the square of the radius of the Earth.
correct
Explanation:
The acceleration
g
∝
1
r
2
of gravity would
increase if the Earth shrank but its mass
stayed the same.
003
(part 1 of 2) 10.0 points
The two blocks are connected by a light
string that passes over a frictionless pulley
with a negligible mass.
The block of mass
m
1
lies on a rough horizontal surface with a
constant coefficient of kinetic friction
μ
. This
block is connected to a spring with spring
constant
k
. The second block has a mass
m
2
.
The system is released from rest when the
spring is unstretched, and
m
2
falls a distance
h
before it reaches the lowest point.
Note:
When
m
2
is at the lowest point, its
velocity is zero.
m
1
m
2
k
m
1
m
2
h
h
μ
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Version 014 – TEST02 – TSOI – (58160)
2
Consider the moment when
m
2
has de
scended by a distance
s
, where
s
is less than
h
.
At this moment the sum of the kinetic energy
for the two blocks
K
is given by
1.
K
= (
m
1
+
m
2
)
g s
+
1
2
k s
2

μ m
1
g s .
2.
K
= (
m
1
+
m
2
)
g s

1
2
k s
2
+
μ m
1
g s .
3.
K
= (
m
1
+
m
2
)
g s

1
2
k s
2

μ m
1
g s .
4.
K
=

(
m
1
+
m
2
)
g s
+
1
2
k s
2
+
μ m
1
g s .
5.
K
=

m
2
g s
+
1
2
k s
2
+
μ
(
m
1
+
m
2
)
g s .
6.
K
=

m
2
g s
+
1
2
k s
2
+
μ m
1
g s .
7.
K
=
m
2
g s

1
2
k s
2

μ
(
m
1
+
m
2
)
g s .
8.
K
=
m
2
g s

1
2
k s
2

μ m
1
g s .
correct
Explanation:
Basic Concepts:
WorkEnergy Theorem
Spring Potential Energy
Frictional
Force
according
to
the
Work
Energy Theorem
Solution:
W
ext
A
→
B
= (
K
B

K
A
) + (
U
g
B

U
g
A
)
+ (
U
sp
B

U
sp
A
) +
W
dis
A
→
B
For
the
present
case,
the
external
work
W
ext
A
→
B
= 0,
A
corresponds to the initial state
and
B
the state where
m
2
has descended by a
distance
s
. The sum of the kinetic energy of
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 Spring '08
 Turner
 Physics, Energy, Force, Friction, Mass, Potential Energy

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