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le (txl65) – Work and Energy – mohanty – (81440)
1
This printout should have 26 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
A 12 g bullet is accelerated in a ri±e barrel
72
.
1 cm long to a speed oF 780 m
/
s.
Use the workenergy theorem to fnd the
average Force exerted on the bullet while it is
being accelerated.
Correct answer: 5062
.
97 N.
Explanation:
Let :
m
= 12 g = 0
.
012 kg
,
x
= 72
.
1 cm
,
and
v
= 780 m
/
s
.
The kinetic energy oF the bullet beFore it
is fred is 0 J since it is not moving.
At
the moment that it leaves the ri±e barrel, its
kinetic energy is
K
=
1
2
mv
2
=
1
2
(0
.
012 kg) (780 m
/
s)
2
= 3650
.
4 J
.
Since the work done on the bullet is equal to
its change in kinetic energy, this is also the
work that is done on the bullet:
W
=
1
2
mv
2
= 3650
.
4 J
.
Also, the work is given by
W
=
F
avg
x
F
avg
=
W
x
=
3650
.
4 J
0
.
721 m
=
5062
.
97 N
.
002 (part 1 of 2) 10.0 points
A bead slides without Friction around a loop
theloop. The bead is released From a height
h
From the bottom oF the looptheloop which
has a radius
r
.
h
r
Which oF the Following diagrams best rep
resents the kinetic energy oF the bead versus
time?
1.
t
K
2.
t
K
correct
3.
t
K
4.
t
K
5.
t
K
6.
t
K
Explanation:
The bead reaches a maximum kinetic en
ergy each time it reaches the bottom oF the
loop. Consequently, the only curve which fts
this criteria is one which has two equally high
maxima. Also, the kinetic energy must be
zero at
t
= 0.
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View Full Document le (txl65) – Work and Energy – mohanty – (81440)
2
003 (part 2 of 2) 10.0 points
Which of the following could represent the
gravitational potential energy of the bead ver
sus time?
1.
t
U
2.
t
U
3.
t
U
4.
t
U
5.
t
U
6.
t
U
correct
Explanation:
Since the kinetic energy plus potential en
ergy is constant, this curve must be the inverse
of the curve in Part 1.
004
10.0 points
An ore car of mass 35000 kg starts from rest
and rolls downhill on tracks from a mine. At
the end of the tracks, 20 m lower vertically, is
a horizontally situated spring with constant
2
.
1
×
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?
Correct answer: 8
.
0829 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
mg h
=
1
2
mv
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
mv
2
=
1
2
k d
2
.
Thus,
1
2
k d
2
=
mg h,
and solving for
d
we have
d
=
r
2
mg h
k
=
R
2 (35000 kg) (9
.
8 m
/
s
2
) (20 m)
(2
.
1
×
10
5
N
/
m)
=
8
.
0829 m
.
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This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.
 Spring '11
 sdsd

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