wolz (cmw2833) – HW #5 – Antoniewicz – (57420)
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001(part1of2)5.0points
A 2
.
01 kg block is placed on top of a 8
.
17 kg
block.
A horizontal force of
F
= 63 N is
applied to the 8
.
17 kg block, and the 2
.
01 kg
block is tied to the wall.
The coefficient of
kinetic friction between all moving surfaces
is 0
.
174.
There is friction both between the
masses and between the 8
.
17 kg block and the
ground.
The acceleration of gravity is 9
.
8 m
/
s
2
.
2
.
01 kg
8
.
17 kg
μ
= 0
.
174
μ
= 0
.
174
F
T
Determine the tension
T
in the string.
Correct answer: 3
.
42745 N.
Explanation:
Given :
m
= 2
.
01 kg
,
M
= 8
.
17 kg
,
m
+
M
= 10
.
18 kg
,
μ
= 0
.
174
,
and
F
= 63 N
.
BasicConcept:
Newton’s second law.
Solution:
Consider the free body diagram
for the situation.
m
N
1
m g
μ m g
T
M
N
1
=
m g
N
2
M g
μ
(
m
+
M
)
g
μ m g
F
Applying Newton’s second law on block
m
yields
m
:
summationdisplay
F
y
=
N
1

m g
= 0
(1)
m
:
summationdisplay
F
x
=
f
1

T
= 0
,
(2)
where
N
1
is the normal force exerted on block
m
by block
M
and
f
1
is the frictional force
between block
m
and block
M
. Using equa
tion (2) to solve for
T
and noting that from
equation (1),
N
1
=
mg
, we obtain
T
=
f
1
=
μ
N
1
=
μ m g
= (0
.
174) (2
.
01 kg) (9
.
8 m
/
s
2
)
= 3
.
42745 N
.
002(part2of2)5.0points
Determine the magnitude of the acceleration
of the 8
.
17 kg block.
Correct answer: 5
.
1669 m
/
s
2
.
Explanation:
Applying Newton’s second law on block
M
yields
M
:
summationdisplay
F
y
=
N
2
 N
1

M g
= 0
(3)
M
:
summationdisplay
F
x
=
F

f
1

f
2
=
M a ,
(4)
where
N
2
is the normal force on block
M
by
the ground and
f
2
is the frictional force be
tween block
M
and the ground. From equa
tion (3),
N
2
=
M g
+
N
1
=
M g
+
m g
= (
M
+
m
)
g .
Thus
f
2
=
μ
N
2
=
μ
(
M
+
m
)
g
Then solving for
a
from (4), we have
a
=
1
M
(
F

f
1

f
2
)
=
1
M
bracketleftBig
F

μ m g

μ
(
M
+
m
)
g
bracketrightBig
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wolz (cmw2833) – HW #5 – Antoniewicz – (57420)
2
=
1
M
bracketleftBig
F

μ
(2
m
+
M
)
g
bracketrightBig
=
1
8
.
17 kg
braceleftBig
63 N

(0
.
174)
bracketleftBig
2 (2
.
01 kg) + (8
.
17 kg)
bracketrightBig
×
(9
.
8 m
/
s
2
)
bracerightBig
= 5
.
1669 m
/
s
2
.
003(part1of2)5.0points
A small metal ball is suspended from the ceil
ing by a thread of negligible mass.
The ball
is then set in motion in a horizontal circle so
that the thread describes a cone.
The acceleration of gravity is 9
.
8 m
/
s
2
.
v
9
.
8 m
/
s
2
2
.
3 m
6
.
7 kg
20
◦
What is the speed of the ball when it is in
circular motion?
Correct answer: 1
.
67508 m
/
s.
Explanation:
Let :
ℓ
= 2
.
3 m
,
θ
= 20
◦
,
g
= 9
.
8 m
/
s
2
,
and
m
= 6
.
7 kg
.
Use the free body diagram below.
T
m g
θ
The tension on the string can be decom
posed into a vertical component which bal
ances the weight of the ball and a horizontal
component which causes the centripetal ac
celeration,
a
centrip
that keeps the ball on its
horizontal circular path at radius
r
=
ℓ
sin
θ
.
If
T
is the magnitude of the tension in the
string, then
T
vertical
=
T
cos
θ
=
m g
(1)
and
T
horiz
=
m a
centrip
or
T
sin
θ
=
m v
2
ball
ℓ
sin
θ
.
(2)
Solving (1) for
T
yields
T
=
m g
cos
θ
(3)
and substituting (3) into (2) gives
m g
tan
θ
=
m v
2
ball
ℓ
sin
θ
.
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 Spring '08
 Turner
 Physics, Force, Friction, Mass, Correct Answer, mg F

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