§
6 Problem 12 (
a
):
Suppose a person walks about 2 or 3 mph, which is about 1 m/s.
Then the energy is
1
2
mv
2
= 38 J. For a good running speed take a 6 minute mile, which is 10 mph, or about 4.4 m/s. This
corresponds to a kinetic energy of about 730 J.
(
b
): The mass of the electron is given on the inside back cover as
m
= 9
.
109
×
10

31
kg.
As the speed is
v
= 2
.
19
×
10
6
m
/
s, the kinetic energy is
K
=
1
2
mv
2
= 2
.
18
×
10

18
J
.
(
c
): Suppose your shoulder is about 1.5 m above the floor. The velocity at the end is given by
v
2
f
=
v
2
0
+ 2
gd
⇒
v
f
= 5
.
4 m
/
s
.
Then the kinetic energy is about 15 J.
(
d
): We have
K
=
1
2
mv
2
⇒
v
=
radicalbigg
2
K
m
=
radicalBigg
2(100 J)
30 kg
= 2
.
6 m
/
s = 5
.
8 mi
/
hr
.
Yes, this is reasonable.
§
6 Problem 17: The total force perpendicular to the slope is
N

mg
cos
α
= 0
⇒
N
=
mg
cos
α.
The total force parallel to the slope is

F
f

mg
sin
α.
Knowing that
F
f
=
μ
k
N
, the total force
F
bardbl
parallel becomes
F
bardbl
=

μ
k
mg
cos
α

mg
sin
α.
The total distance the box slides is
d
=
h/
sin
α
. The total work done by the parallel forces is
W
=
F
bardbl
d
=

mgh
μ
k
cos
α
+ sin
α
sin
α
.
This is the change in the kinetic energy. As the box just barely reaches the stranded skier, the final velocity
is 0. Thus
W
=
K.E.
f

K.E.
0
=

1
2
mv
2
0
=

mgh
μ
k
cos
α
+ sin
α
sin
α
,
so
v
0
=
radicalbigg
2
gh
(
μ
k
cos
α
+ sin
α
)
sin
α
.
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 Summer '08
 Chamberlin
 Energy, Force, Kinetic Energy, Potential Energy, per capita, 0.40 m

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