C
HAPTER
5S
OLUTION FOR
P
ROBLEM
13
(a) The freebody diagram is shown in Fig. 5–18 of the text. Since the acceleration of the
block is zero, the components of the Newton’s second law equation yield
T
−
mg
sin
θ
= 0 and
F
N
−
mg
cos
θ
= 0. Solve the first equation for the tension force of the string:
T
=
mg
sin
θ
=
(8
.
5 kg)(9
.
8m
/
s
2
) sin 30
◦
=42N
.
(b) Solve the second equation for
F
N
:
F
N
=
mg
cos
θ
=(8
.
5 kg)(9
.
/
s
2
) cos 30
◦
=72N
.
(c) When the string is cut it no longer exerts a force on the block and the block accelerates. The
x
component of the second law becomes
−
mg
sin
θ
=
ma
,so
a
=
−
g
sin
θ
=
−
(9
.
/
s
2
) sin 30
◦
=
−
4
.
9m
/
s
2
. The negative sign indicates the acceleration is down the plane.
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View Full DocumentC
HAPTER
5S
OLUTION FOR
P
ROBLEM
29
The freebody diagram is shown at the right.
n
F
N
is the normal force
of the plane on the block and
m
n
g
is the force of gravity on the block.
Take the positive
x
axis to be down the plane, in the direction of the
acceleration, and the positive
y
axis to be in the direction of the normal
force. The
x
component of Newton’s second law is then
mg
sin
θ
=
ma
, so the acceleration is
a
=
g
sin
θ
.
(a) Place the origin at the bottom of the plane. The equations for
motion along the
x
axis are
x
=
v
0
t
+
1
2
at
2
and
v
=
v
0
+
at
. The block
stops when
v
=0
.
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F
N
mg
θ
According to the second equation, this is at the time
t
=
−
v
0
/a
. The coordinate when it stops is
x
=
v
0
w
−
v
0
a
W
+
1
2
a
w
−
v
0
a
W
2
=
−
1
2
v
2
0
a
=
−
1
2
v
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 Spring '09
 FOBAIR
 Force, 2 g, 3.6 m/s, 1.4 m/s, 2.3 kg, 1.18 m

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