hinojosa (jlh3938) – oldhomework 16 – Turner – (58185)
1
This
printout
should
have
16
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
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.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Physics, Potential Energy, Work, Correct Answer, Hinojosa

Click to edit the document details