61
W
ORK AND
K
INETIC
E
NERGY
6.1.
IDENTIFY:
Apply Eq.(6.2).
SET UP:
The bucket rises slowly, so the tension in the rope may be taken to be the bucket’s weight.
EXECUTE:
(a)
2
(6.75 kg) (9.80 m/s )(4.00 m)
265 J.
WF
sm
g
s
==
=
=
(b)
Gravity is directed opposite to the direction of the bucket’s motion, so Eq.(6.2) gives the negative of the result of
part (a), or
265 J
−
.
(c)
The total work done on the bucket is zero.
EVALUATE:
When the force is in the direction of the displacement, the force does positive work. When the force is
directed opposite to the displacement, the force does negative work.
6.2.
IDENTIFY:
In each case the forces are constant and the displacement is along a straight line, so
cos
s
φ
=
.
SET UP:
In part (a), when the cable pulls horizontally
0
=
°
and when it pulls at 35.0
°
above the horizontal
35.0
=
°
. In part (b), if the cable pulls horizontally
180
=
°
. If the cable pulls on the car at 35.0
°
above the
horizontal it pulls on the truck at 35.0
°
below the horizontal and
145.0
=
°
. For the gravity force
90
=
°
, since the
force is vertical and the displacement is horizontal.
EXECUTE:
(a)
When the cable is horizontal,
36
(850 N)(5.00 10 m)cos0
4.25 10 J
W
=×=
×
°
. When the cable is
35.0
°
above the horizontal,
(850 N)(5.00 10 m)cos35.0
3.48 10 J
W
=×
=
×
°
.
(b)
cos180
cos0
=−
°°
and cos145.0
cos35.0
, so the answers are
6
4.26 10 J
−×
and
6
.
(c)
Since cos
cos90
0
°
,
0
W
=
in both cases.
EVALUATE:
If the car and truck are taken together as the system, the tension in the cable does no net work.
6.3.
IDENTIFY:
Each force can be used in the relation
(c
o
s)
s F
s
&
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
.
Ff
n
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
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
180
=
°
and the work of friction is:
(
)
[ ]
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
+
+
=
+
+
+
−
=
EVALUATE:
The net work done on the crate is zero because the two contributing forces,
worker
and
,
f
FF
are equal in
magnitude and opposite in direction.
6.4.
IDENTIFY:
The forces are constant so Eq.(6.2) can be used to calculate the work. Constant speed implies
0.
a
=
We
must use
m
=
∑
F
a
G
G
applied to the crate to find the forces acting on it.
6
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Chapter 6
(a) SET UP:
The freebody diagram for the crate is given in Figure 6.4.
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 Spring '11
 Shaefer
 Force, Kinetic Energy, Work, K1, Wtot

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