7.3.IDENTIFY:
Use the freebody diagram for the bag and Newton's first law to find the force the worker
applies. Since the bag starts and ends at rest,
21
0
KK
−
=
and
tot
0
W
=
.
SET UP:
A sketch showing the initial and final positions of the bag is given in Figure 7.3a.
2.0 m
sin
3.5 m
φ
=
and
. The freebody diagram is given in Figure 7.3b.
34.85
=
°
F
G
is the horizontal force
applied by the worker. In the calculation of
take
grav
U
y
+
upward and
0
y
=
at the initial position of the
bag.
EXECUTE:
(a)
gives
0
y
F
=
∑
cos
Tm
g
=
and
0
x
F
=
∑
gives
sin
FT
=
. Combining these
equations to eliminate
T
gives
.
2
tan
(120 kg)(9.80 m/s )tan34.85
820 N
Fm
g
==
=
°
(b)
(i) The tension in the rope is radial and the displacement is tangential so there is no component of
T
in the direction of the displacement during the motion and the tension in the rope does no work. (ii)
so
.
tot
0
W
=
2
worker
grav
grav,2
grav,1
2
1
(
)
(120 kg)(9.80 m/s )(0.6277 m)
740 J
WW
U
U
m
g
y
y
=−
=
−
=
−
=
=
EVALUATE:
The force applied by the worker varies during the motion of the bag and it would be
difficult to calculate
directly.
worker
W
Figure 7.3
7.5.
IDENTIFY
and
SET UP:
Use energy methods.
(a)
1
1
other
2
2
.
K
UW
KU
++
= +
Solve for
2
K
and then use
2
1
2
2
2
K
mv
=
to obtain
2
.
v
other
0
W
=
(The only force on the
ball while it is in the air is gravity.)
2
1
11
2
;
K
mv
=
2
1
22
2
K
mv
=
,
Um
g
y
=
1
22.0 m
y
=
0,
g
y
=
=
since
2
0
y
=
for our choice of coordinates.
Figure 7.5
EXECUTE:
mv
mgy
mv
+=
2
2
1
2
(12.0 m/s)
2(9.80 m/s )(22.0 m)
24.0 m/s
vvg
y
=+
=
+
=
EVALUATE:
The projection angle of
doesn’t enter into the calculation. The kinetic energy
depends only on the magnitude of the velocity; it is independent of the direction of the velocity.
53.1
°
(b)
Nothing changes in the calculation. The expression derived in part (a) for
is independent of the
angle, so
the same as in part (a).
2
v
2
24.0 m/s,
v
=
(c)
The ball travels a shorter distance in part (b), so in that case air resistance will have
less effect.
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View Full Document7.9.IDENTIFY:
tot
B
A
WK
K
=−
. The forces on the rock are gravity, the normal force and friction.
SET UP:
Let
at point
B
and let
0
y
=
y
+
be upward.
. The work done by friction is
negative;
W
.
. The freebody diagram for the rock at point
B
is given in Figure 7.9.
The acceleration of the rock at this point is
, upward.
0.50 m
A
yR
==
0.22 J
f
0
A
K
=
R
2
rad
/
av
=
EXECUTE:
(a)
(i) The normal force is perpendicular to the displacement and does zero work.
(ii)
.
2
grav
grav,
grav,
(0.20 kg)(9.80 m/s )(0.50 m)
0.98 J
ABA
WU
U
m
g
y
=−=
=
=
(b)
.
tot
grav
0
( 0.22 J)
0.98 J
0.76 J
nf
WW
W
W
=++ =
+
−
+
=
tot
B
A
K
gives
2
1
tot
2
B
mv
W
=
.
tot
22
(
0
.
7
6
J
)
2.8 m/s
0.20 kg
B
W
v
m
=
.
(c)
Gravity is constant and equal to
mg
.
n
is not constant; it is zero at
A
and not zero at
B
. Therefore,
kk
f
n
μ
=
is also not constant.
(d)
y
y
Fm
a
=
∑
applied to Figure 7.9 gives
.
rad
nm
gm
a
−=
2
[2.8 m/s]
(0.20 kg) 9.80 m/s
5.1 N
0.50 m
v
nmg
R
⎛⎞
⎛
⎞
=+
=
+
=
⎜⎟
⎜
⎟
⎝⎠
⎝
⎠
.
EVALUATE:
In the absence of friction, the speed of the rock at point
B
would be
23
.
1
m
gR
=
/
s
. As
the rock slides through point
B
, the normal force is greater than the weight
2.0 N
mg
=
of the rock.
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 Fall '08
 Kiel
 Energy, Force, Friction, Potential Energy, Work, m/s, Wother

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