min (mm64262) – Homework 3 – lee – (GEDB00864)
1
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001
10.0points
The sketch shows a coin at the edge of a
turntable. The weight of the coin is shown by
the vector
W
.
Two other forces act on the
coin: the normal force and a force of friction
that prevents it from sliding off the edge.
ω
W
Draw the two remaining force fectors.
1.
f
N
2.
f
N
correct
3.
f
N
4.
None of these graphs is correct.
5.
f
N
6.
f
N
7.
f
N
8.
f
N
9.
f
N
Explanation:
f
W
N
The friction force must point inward in or
der to keep the coin moving in a circle (other
wise it will fly off the edge of the disk), and the
normal force counteracts the force of gravity.
002
10.0points
A crate is pulled by a force (parallel to the
incline) up a rough incline. The crate has an
initial speed shown in the figure below. The
crate is pulled a distance of 5
.
21 m on the
incline by a 150 N force.
The acceleration of gravity is 9
.
8 m
/
s
2
.
13 kg
μ
= 0
.
287
150 N
1
.
29 m
/
s
21
◦
What is the change in kinetic energy of the
crate?
Correct answer: 365
.
787 J.
Explanation:
Let :
F
= 150 N
,
d
= 5
.
21 m
,
θ
= 21
◦
,
m
= 13 kg
,
g
= 9
.
8 m
/
s
2
,
μ
= 0
.
287
,
and
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min (mm64262) – Homework 3 – lee – (GEDB00864)
2
v
= 1
.
29 m
/
s
.
F
μ N
N
m g
v
θ
The workenergy theorem with nonconser
vative forces reads
W
fric
+
W
appl
+
W
gravity
= Δ
K
To find the work done by friction we need the
normal force on the block from Newton’s law
summationdisplay
F
y
=
N 
m g
cos
θ
= 0
⇒ N
=
m g
cos
θ .
Thus
W
fric
=

μ m g d
cos
θ
=

(0
.
287) (13 kg) (9
.
8 m
/
s
2
)
×
(5
.
21 m) cos 21
◦
=

177
.
845 J
.
The work due to the applied force is
W
appl
=
F d
= (150 N) (5
.
21 m)
= 781
.
5 J
,
and the work due to gravity is
W
grav
=

m g d
sin
θ
=

(13 kg) (9
.
8 m
/
s
2
)
×
(5
.
21 m) sin 21
◦
=

237
.
868 J
,
so that
Δ
K
=
W
fric
+
W
appl
+
W
grav
= (

177
.
845 J) + (781
.
5 J)
+ (

237
.
868 J)
= 365
.
787 J
.
003(part1of2)10.0points
A spring has a force constant of 819 N
/
m and
an unstretched length of 6 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
1 kg disk moving in uniform circular motion
on the table, which stretches the spring by
2 cm.
Note:
Friction is negligible.
819 N
/
m
1 kg
8 cm
What is the centripetal force
F
c
on the disk?
Correct answer: 16
.
38 N.
Explanation:
Let :
r
= 6 cm = 0
.
06 m
,
Δ
r
= 2 cm = 0
.
02 m
,
m
= 1 kg
,
and
k
= 819 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
= (819 N
/
m) (0
.
02 m)
=
16
.
38 N
.
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 Spring '11
 LEE
 Force, Sin, tan θ

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