hinojosa (jlh3938) – oldhomework 16 – Turner – (58185)
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001
(part 1 of 3) 10.0 points
A 2
.
7 kg particle moves along the
x
axis under
the influence of a single conservative force.
If the work done on the particle is 106 J as
it moves from
x
1
= 1 m to
x
2
= 4 m, find the
change in its kinetic energy.
Correct answer: 106 J.
Explanation:
The change in the kinetic energy is equal to
the work done:
Δ
K
=
W
=
106 J
.
002
(part 2 of 3) 10.0 points
Find the change in its potential energy.
Correct answer:

106 J.
Explanation:
Since
Δ
K
+ Δ
U
= 0
,
then
Δ
U
=

Δ
K
=

106 J
.
003
(part 3 of 3) 10.0 points
Find its speed at
x
2
= 4 m if it starts from
rest.
Correct answer: 8
.
86107 m
/
s.
Explanation:
Using
K
=
m
(
v
2
end

v
2
init
)
2
we obtain
v
=
radicalbigg
2Δ
K
m
=
radicalBigg
2(106 J)
2
.
7 kg
=
8
.
86107 m
/
s
.
004
10.0 points
A block of mass
m
slides on a horizontal
frictionless table with an initial speed
v
0
.
It
then compresses a spring of force constant
k
and is brought to rest.
The acceleration of gravity is 9
.
8 m
/
s
2
.
v
m
k
m
μ
= 0
How much is the spring compressed
x
from
its natural length?
1.
x
=
v
2
0
2
m
2.
x
=
v
0
m k
g
3.
x
=
v
0
m
k g
4.
x
=
v
0
radicalbigg
m g
k
5.
x
=
v
0
radicalbigg
m
k
correct
6.
x
=
v
0
radicalbigg
k
m
7.
x
=
v
0
m g
k
8.
x
=
v
0
radicalBigg
k
m g
9.
x
=
v
2
0
2
g
10.
x
=
v
0
k
g m
Explanation:
Total energy is conserved (no friction). The
spring is compressed by a distance
x
from its
natural length, so
1
2
m v
2
0
=
E
i
=
E
f
=
1
2
k x
2
,
or
x
2
=
m
k
v
2
0
,
therefore
x
=
v
0
radicalbigg
m
k
.
Anyone who checks to see if the units are
correct should get this problem correct.
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hinojosa (jlh3938) – oldhomework 16 – Turner – (58185)
2
005
10.0 points
A bead slides without friction around a loop
theloop. The bead is released from a height
of 12 m from the bottom of the looptheloop
which has a radius 4 m.
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 Fall '08
 Turner
 Potential Energy, Work, Correct Answer, Hinojosa

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