gutierrez (ig3472) – HW5 – ditmire – (58216)
1
This printout should have 26 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 2) 10.0 points
A spring has a Force constant oF 311 N
/
m and
an unstretched length oF 7 cm. One end is
attached to a post that is Free to rotate in the
center oF a smooth table, as shown in the top
view below. The other end is attached to a
9 kg disk moving in uniForm circular motion
on the table, which stretches the spring by
1 cm.
Note:
±riction is negligible.
311 N
/
m
9 kg
8 cm
What is the centripetal Force
F
c
on the disk?
Correct answer: 3
.
11 N.
Explanation:
Let :
r
= 7 cm = 0
.
07 m
,
Δ
r
= 1 cm = 0
.
01 m
,
m
= 9 kg
,
and
k
= 311 N
/
m
.
The centripetal Force is supplied only by
the spring. Given the Force constant and the
extension oF the spring, we can calculate the
Force as
F
c
=
k
Δ
r
= (311 N
/
m) (0
.
01 m)
=
3
.
11 N
.
002
(part 2 oF 2) 10.0 points
What is the work done on the disk by the
spring during one Full circle?
1.
W
= 0
.
00106673 J
2.
W
= 0 J
correct
3.
W
= 1
.
36785 J
4.
W
= 0
.
2177 J
5.
W
= 0
.
683925 J
Explanation:
Since the Force is always perpendicular to
the movement oF the disk, the work done by
the spring is
zero
.
003
(part 1 oF 3) 10.0 points
A 1270 kg car accelerates uniFormly From rest
to 14
.
2 m
/
s in 4
.
02 s.
±ind the work done on the car in this time
interval.
Correct answer: 128
.
041 kJ.
Explanation:
Given :
m
= 1270 kg
,
v
f
= 14
.
2 m
/
s
,
and
Δ
t
= 4
.
02 s
.
W
= Δ
K
=
1
2
mv
2
f

0
=
1
2
(1270 kg) (14
.
2 m
/
s)
2
·
1 kJ
1000 J
=
128
.
041 kJ
.
004
(part 2 oF 3) 10.0 points
±ind the average power delivered by the en
gine in this time interval.
Correct answer: 42
.
6958 hp.
Explanation:
P
=
W
Δ
t
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documentgutierrez (ig3472) – HW5 – ditmire – (58216)
2
=
p
128
.
041 kJ
4
.
02 s
Pp
1000 J
kJ
1 hp
746 W
P
=
42
.
6958 hp
.
005
(part 3 of 3) 10.0 points
Find the instantaneous power delivered by the
engine at
t
= 1
.
84 s.
Correct answer: 39
.
0847 hp.
Explanation:
Given :
Δ
t
1
= 1
.
84 s
In the ±rst 4
.
02 s
a
=
Δ
v
Δ
t
=
v
f

v
i
Δ
t
=
v
f
Δ
t
since
v
i
= 0, so the net force is
F
=
ma
=
mv
f
Δ
t
.
At
t
= 1
.
84 s
,
the instantaneous velocity is
v
=
v
i
+
a
Δ
t
1
=
a
Δ
t
1
=
v
f
Δ
t
Δ
t
1
since
v
i
= 0
,
and the power output is
P
=
F v
=
f
Δ
t
1
·
±
v
f
Δ
t
Δ
t
1
²
=
m
±
v
f
Δ
t
²
2
Δ
t
1
= (1270 kg)
p
14
.
2 m
/
s
4
.
02 s
P
2
·
(1
.
84 s)
p
1 hp
746 W
P
=
39
.
0847 hp
.
006
(part 1 of 5) 10.0 points
A crate is pulled up a rough incline.
The
pulling force is parallel to the incline. The
crate is pulled a distance of 5
.
04 m.
The acceleration of gravity is 9
.
8 m
/
s
2
.
9
.
69 kg
μ
= 0
281
169 N
1
69 m
/
s
28
.
6
◦
What is the magnitude of the work is done
by the gravitational force?
Correct answer: 229
.
106 J.
Explanation:
Let :
d
= 5
.
04 m
,
θ
= 28
.
6
◦
,
m
= 9
.
69 kg
,
g
= 9
.
8 m
/
s
2
,
μ
= 0
.
281
,
and
v
= 1
.
69 m
/
s
.
F
μ
N
N
mg
v
θ
The gravitational force directed down the
plane is
F
grav
=

sin
θ
, and the motion
is directed up the plane, so the work by the
gravity is
W
grav
=

mg d
sin
θ
=

(9
.
69 kg)
g
(5
.
04 m) sin28
.
6
◦
=

229
.
106 J

W
grav

= 229
.
106 J
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Physics, Energy, Force, Kinetic Energy, Potential Energy, Correct Answer

Click to edit the document details