7-20
Chapter 7
7.61.
IDENTIFY
and
SET UP:
There are two situations to compare: stepping off a platform and sliding down a pole.
Apply the work-energy theorem to each.
(a)
EXECUTE:
Speed at ground if steps off platform at height
h
:
1
1
other
2
2
K
UW
KU
++
= +
2
1
2
2
,
mgh
mv
=
so
2
2
2
vg
h
=
Motion from top to bottom of pole: (take
0
y
=
at bottom)
1
1
other
2
2
K
2
1
2
2
mgd
fd
mv
−=
Use
2
2
2
h
=
and get
mgd
fd
mgh
()
f
dm
g
dh
=−
/
(
1
/
)
f
mg d
h d
mg
h d
=
−
EVALUATE:
For
hd
=
this gives
0
f
=
as it should (friction has no effect).
For
0,
h
=
2
0
v
=
(no motion). The equation for
f
gives
f
mg
=
in this special case. When
f
mg
=
the forces on
him cancel and he doesn&t accelerate down the pole, which agrees with
2
0.
v
=
(b)
EXECUTE:
2
(1
/ )
(75 kg)(9.80 m/s )(1 1.0 m/2.5 m)
441 N.
fm
g h
d
=−=
−
=
(c)
Take
0
y
=
at bottom of pole, so
1
yd
=
and
2
.
yy
=
1
1
other
2
2
K
2
1
2
0(
)
mgd
f d
y
mv
mgy
+−−
=
+
2
1
2
mv
mg d
y
f d
y
−
−
Using
/ )
f
mg
h d
gives
2
1
2
()(
1
/
)
mv
mg d
y
mg
h d d
y
−
−
−
2
1
2
(/)
(
)
mv
mg h d d
y
and
2(
1 /)
h
y
d
EVALUATE:
This gives the correct results for
0
y
=
and for
.
=
7.62.
IDENTIFY:
Apply Eq.(7.14) to each stage of the motion.
SET UP:
Let
0
y
=
at the bottom of the slope. In part (a),
other
W
is the work done by friction. In part (b),
other
W
is
the work done by friction and the air resistance force. In part (c),
other
W
is the work done by the force exerted by the
snowdrift.
EXECUTE:
(a)
The skier&s kinetic energy at the bottom can be found from the potential energy at the top minus
the work done by friction,
1
(60.0 kg)(9.8 N/kg)(65.0 m) 10,500 J,
f
Km
g
hW
=
−
or
1
38,200 J 10,500 J
27,720 J
K
. Then
1
1
2
2(27,720 J)
30.4 m/s
60 kg
K
v
m
==
=
.
(b)
2
1
air
k
air
(
)
27,720 J
(
).
f
K
KWW
m
g
df
d
μ
=− + =
−
+
2
27,720 J [(0.2)(588 N)(82 m)
(160 N)(82 m)]
K
+
or
2
27,720 J
22,763 J
4957 J
K
. Then,
2
2
2(4957 J)
12.9 m/s
60 kg
K
v
m
=
(c)
Use the Work-Energy Theorem to find the force.
,
WK
= Δ
/
(4957 J) (2.5 m)
2000 N
FK
d
=
.
EVALUATE:
In each case,
other
W
is negative and removes mechanical energy from the system.
7.63.
IDENTIFY
and
SET UP:
First apply
m
=
∑
F
a
!
!
to the skier.
Find the angle
α
where the normal force becomes zero, in terms of the speed
v
2
at this point. Then apply the
work-energy theorem to the motion of the skier to obtain another equation that relates
v
2
and
.
Solve these two
equations for
.
Let point 2 be where the skier loses contact
with the snowball, as sketched in Figure 7.63a
Loses contact implies
0.
n
→
1
,
yR
=
2
cos
=
Figure 7.63a