portillo (gdp347) – homework 35 – Turner – (58185)
1
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001
(part 1 of 2) 10.0 points
A 1
.
47 kg block at rest on a tabletop is at
tached to a horizontal spring having constant
19
.
5 N
/
m. The spring is initially unstretched.
A constant 26
.
3 N horizontal force is applied
to the object causing the spring to stretch.
26
.
3 N
1
.
47 kg
19
.
5 N
/
m
Find the speed of the block after it has
moved 0
.
209 m from equilibrium if the sur
face between block and tabletop is friction
less. The acceleration of gravity is 9
.
8 m
/
s
2
.
Correct answer: 2
.
62661 m
/
s.
Explanation:
Let :
m
= 1
.
47 kg
,
x
f
= 0
.
209 m
,
k
= 19
.
5 N
/
m
,
and
F
= 26
.
3 N
.
Applying the work kinetic energy theorem,
F x
f
=
1
2
m v
2
f
+
1
2
k x
2
f
m v
2
f
= 2
F x
f

k x
2
f
v
2
f
=
2
F x
f

k x
2
f
m
=
2 (26
.
3 N) (0
.
209 m)
1
.
47 kg

(19
.
5 N
/
m) (0
.
209 m)
2
1
.
47 kg
= 6
.
89906 m
2
/
s
2
v
f
=
radicalBig
6
.
89906 m
2
/
s
2
=
2
.
62661 m
/
s
.
002
(part 2 of 2) 10.0 points
Find the speed of the block after it has moved
0
.
209 m from equilibrium if the coefficient of
kinetic friction between block and tabletop is
0
.
142.
Correct answer: 2
.
51344 m
/
s.
Explanation:
Let :
μ
k
= 0
.
142
.
Applying the work kinetic energy theorem,
F x
f

μ
k
m g x
f
=
1
2
m v
2
f
+
1
2
k x
2
f
m v
2
f
= 2
F x
f

2
μ
k
m g x
f

k x
2
f
v
2
f
=
2
F x
f

2
μ
k
m g x
f

k x
2
f
m
=
2 (26
.
3 N) (0
.
209 m)
1
.
47 kg

2 (0
.
142)
(
9
.
8 m
/
s
2
)
(0
.
209 m)

(19
.
5 N
/
m) (0
.
209 m)
2
1
.
47 kg
= 6
.
31737 m
2
/
s
2
v
f
=
radicalBig
6
.
31737 m
2
/
s
2
=
2
.
51344 m
/
s
.
003
10.0 points
A 15 kg block is connected to a 25 kg block
by a light string that passes over a frictionless
pulley having negligible mass. The 33
◦
incline
is smooth (frictionless).
The acceleration of
gravity is 9
.
8 m
/
s
2
.
The 25 kg block is connected to a light
spring of force constant 248 N
/
m.
The spring is unstretched when the 25 kg
block is 13 cm above the floor (as shown in
the figure).
The 15 kg block is pulled 18 cm down the
incline (so that the 25 kg block is 31 cm above
the floor) and is released from rest.
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portillo (gdp347) – homework 35 – Turner – (58185)
2
25 kg
T
1
15 kg
μ
= 0
T
2
18 cm
33
◦
248 N
/
m
13 cm
31 cm
Find the speed of the 15 kg block when the
25 kg block is 13 cm above the floor (that is,
when the spring is unstretched).
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 Spring '10
 McCord
 Energy, Force, Mass, Simple Harmonic Motion

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