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6.3.IDENTIFY:
Each force can be used in the relation
(c
o
s)
WF
s F
s
φ
==
&
for parts (b) through (d). For part (e),
apply the net work relation as
net
worker
grav
.
nf
WW
W
W
W
=+
+
+
SET UP:
In order to move the crate at constant velocity, the worker must apply a force that equals the
force of friction,
worker
k
k
.
Ffn
μ
EXECUTE:
(a)
The magnitude of the force the worker must apply is:
()
(
)
2
worker
k
k
k
0.25 30.0 kg 9.80 m/s
74 N
Ff
n
m
g
μμ
== =
=
=
(b)
Since the force applied by the worker is horizontal and in the direction of the displacement,
0
=
°
and the work is:
(
)(
)
[ ]( )
worker
worker
cos
74 N cos0
4.5 m
333 J
s
=
°
+
(c)
Friction acts in the direction opposite of motion, thus
and the work of friction is:
180
=
°
( )
(
)
[ ]( )
k
cos
74 N cos180
4.5 m
333 J
f
Wf
s
=
−
°
(d)
Both gravity and the normal force act perpendicular to the direction of displacement. Thus, neither
force does any work on the crate and
grav
0.0 J.
n
(e)
Substituting into the net work relation, the net work done on the crate is:
net
worker
grav
333 J
0.0 J
0.0 J
333 J
0.0 J
W
W
W
+
+
=
+
+
+
−
=
E
VALUATE
:
The net work done on the crate is zero because the two contributing forces,
are equal in magnitude and opposite in direction.
worker
and
,
f
FF
6.4.IDENTIFY:
The forces are constant so Eq.(6.2) can be used to calculate the work. Constant speed implies
We must use
0.
a
=
m
=
∑
F
a
G
G
applied to the crate to find the forces acting on it.
(a) SET UP:
The freebody diagram for the crate is given in Figure 6.4.
EXECUTE:
y
y
Fm
a
=
∑
sin30
0
nm
gF
−
−°
=
sin30
=
+°
kk
k
k
sin30
fn
m
g
F
=
°
Figure 6.4
x
x
a
=
∑
k
cos30
0
°−
=
cos30
sin30
0
g
−
° =
F
2
k
k
0.25(30.0 kg)(9.80 m/s )
99.2 N
cos30
sin30
cos30
(0.25)sin30
mg
F
°
°
=
(b)
( cos )
(99.2 N)(cos30 )(4.5 m)
387 J
F
s
°
=
( cos30
F
°
is the horizontal component of
;
F
G
the work done by
F
G
is the displacement times the
component of
F
G
in the direction of the displacement.)
(c)
We have an expression for
k
f
from part (a):
2
(
sin30 )
(0.250)[(30.0 kg)(9.80 m/s )
(99.2 N)(sin30 )]
85.9 N
fm
g
F
°
=
+
°
=
180
=°
since
k
f
is opposite to the displacement. Thus
k
(
cos )
(85.9 N)(cos180 )(4.5 m)
387 J
f
s
°
=
−
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View Full Document(d)
The normal force is perpendicular to the displacement so
90
φ
=
°
and
0.
n
W
=
The gravity force
(the weight) is perpendicular to the displacement so
90
=
°
and
0
w
W
=
(e)
tot
387 J
( 387 J)
0
Ffnw
WW
W
W
W
=+++=
+
+
−
=
E
VALUATE
:
Forces with a component in the direction of the displacement do positive
work, forces opposite to the displacement do negative work and forces perpendicular to
the displacement do zero work. The total work, obtained as the sum of the work done by
each force, equals the work done by the net force. In this problem,
net
0
F
=
since
and
which agrees with the sum calculated in part (e).
0
a
=
tot
0,
W
=
6.17.IDENTIFY
and
SET UP:
Apply Eq.(6.6) to the box. Let point 1 be at the bottom of the incline and let point
2 be at the skier. Work is done by gravity and by friction. Solve for
1
K
and from that obtain the
required initial speed.
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 Fall '08
 Kiel
 Force, Work

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