71
P
OTENTIAL
E
NERGY AND
E
NERGY
C
ONSERVATION
7.1.
I
DENTIFY
:
grav
U
mgy
=
so
grav
2
1
(
)
U
mg y
y
Δ
=
−
S
ET UP
:
y
+
is upward.
E
XECUTE
:
(a)
2
5
(75 kg)(9.80 m/s
)(2400 m
1500 m)
6.6
10
J
U
Δ
=
−
= +
×
(b)
2
5
(75 kg)(9.80 m/s
)(1350 m
2400 m)
7.7
10
J
U
Δ
=
−
= −
×
E
VALUATE
:
grav
U
increases when the altitude of the object increases.
7.2.
I
DENTIFY
:
Apply
m
=
∑
F
a
G
G
to the sack to find the force.
cos
W
Fs
φ
=
.
S
ET UP
:
The lifting force acts in the same direction as the sack’s motion, so
0
φ
=
°
E
XECUTE
:
(a)
For constant speed, the net force is zero, so the required force is the sack’s weight,
2
(5.00 kg)(9.80 m/s
)
49.0 N.
=
(b)
(49.0 N) (15.0 m)
735 J
W
=
=
. This work becomes potential energy.
E
VALUATE
:
The results are independent of the speed.
7.3.
I
DENTIFY
:
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,
2
1
0
K
K
−
=
and
tot
0
W
=
.
S
ET 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
34.85
φ
=
°
. The freebody diagram is given in Figure 7.3b.
F
G
is the horizontal force applied by the worker. In the
calculation of
grav
U
take
y
+
upward and
0
y
=
at the initial position of the bag.
E
XECUTE
:
(a)
0
y
F
=
∑
gives
cos
T
mg
φ
=
and
0
x
F
=
∑
gives
sin
F
T
φ
=
. Combining these equations to
eliminate
T
gives
2
tan
(120 kg)(9.80 m/s
)tan34.85
820 N
F
mg
φ
=
=
=
°
.
(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)
tot
0
W
=
so
2
worker
grav
grav,2
grav,1
2
1
(
)
(120 kg)(9.80 m/s
)(0.6277 m)
740 J
W
W
U
U
mg y
y
= −
=
−
=
−
=
=
.
E
VALUATE
:
The force applied by the worker varies during the motion of the bag and it would be difficult to
calculate
worker
W
directly.
Figure 7.3
7.4.
I
DENTIFY
:
Only gravity does work on him from the point where he has just left the board until just before he
enters the water, so Eq.(7.4) applies.
S
ET UP
:
Let point 1 be just after he leaves the board and point 2 be just before he enters the water.
y
+
is upward
and
0
y
=
at the water.
7
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72
Chapter 7
E
XECUTE
:
(a)
1
0
K
=
.
2
0
y
=
.
1
3.25 m
y
=
.
1
grav,1
2
grav,2
K
U
K
U
+
=
+
gives
grav,1
2
U
K
=
and
2
1
1
2
2
mgy
mv
=
.
2
2
1
2
2(9.80 m/s
)(3.25 m)
7.98 m/s
v
gy
=
=
=
.
(b)
1
2.50 m/s
v
=
,
2
0
y
=
,
1
3.25 m
y
=
.
1
grav,1
2
K
U
K
+
=
and
2
2
1
1
1
1
2
2
2
mv
mgy
mv
+
=
.
2
2
2
2
1
1
2
(2.50 m/s)
2(9.80 m/s
)(3.25 m)
8.36 m/s
v
v
gy
=
+
=
+
=
.
(c)
1
2.5 m/s
v
=
and
2
8.36 m/s
v
=
, the same as in part (b).
E
VALUATE
:
Kinetic energy depends only on the speed, not on the direction of the velocity.
7.5.
I
DENTIFY
and
S
ET UP
:
Use energy methods.
(a)
1
1
other
2
2
.
K
U
W
K
U
+
+
=
+
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
1
1
2
;
K
mv
=
2
1
2
2
2
K
mv
=
1
1
,
U
mgy
=
1
22.0 m
y
=
2
2
0,
U
mgy
=
=
since
2
0
y
=
for our choice of coordinates.
Figure 7.5
E
XECUTE
:
2
2
1
1
1
1
2
2
2
mv
mgy
mv
+
=
2
2
2
2
1
1
2
(12.0 m/s)
2(9.80 m/s
)(22.0 m)
24.0 m/s
v
v
gy
=
+
=
+
=
E
VALUATE
:
The projection angle of 53.1
°
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.
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 Spring '11
 Shaefer
 Energy, Force, Friction, Potential Energy, Wother

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