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Work and Kinetic Energy
623
6.86.
IDENTIFY:
av
av
PF
v
=
!
. Use
Fm
a
=
to calculate the force.
SET UP:
av
06
.
0
0
m
/
s
3.00 m/s
2
v
+
==
EXECUTE:
Your friend&s average acceleration is
2
0
6.00 m/s
2.00 m/s
3.00 s
vv
a
t
−
=
. Since there are no other
horizontal forces acting, the force you exert on her is given by
2
net
(65.0 kg)(2.00 m/s ) 130 N
a
=
.
av
(130 N)(3.00 m/s)
390 W
P
.
EVALUATE:
We could also use the workenergy theorem:
2
1
21
2
(65.0 kg)(6.00 m/s)
1170 J
WK K
=−=
=
.
av
1170 J
390 W
3.00 s
W
P
t
=
, the same as obtained by our other approach.
6.87.
IDENTIFY:
To lift a mass
m
a height
h
requires work
Wm
g
h
=
. To accelerate mass
m
from rest to speed
v
requires
2
1
2
WK K m
v
.
av
W
P
t
Δ
=
Δ
.
SET UP:
60 s
t
=
EXECUTE:
(a)
25
(800 kg)(9.80 m/s )(14.0 m) 1.10 10 J
=×
(b)
(1/2)(800 kg)(18.0 m/s ) 1.30 10 J.
(c)
55
1.10 10 J 1.30 10 J
3.99 kW.
60 s
×+
×
=
EVALUATE:
Approximately the same amount of work is required to lift the water against gravity as to accelerate it
to its final speed.
6.88.
IDENTIFY:
v
=
!
and
a
=
!
.
SET UP:
From Problem 6.71,
2
23
vtt
αβ
=+
and
2
6
at
α
β
=
+
.
EXECUTE:
222
2
3
(2
6
)(2
3
)
(4
18
18
)
vm
a
β
tt
β
tmt
β
t
β
t
αα
=
+
+
=
+
+
!
.
22
33
(0.96 N/s)
(0.43 N/s )
(0.043 N/s )
Pt
t
t
+
. At
4.00 s,
t
=
the power output is 13.5 W.
EVALUATE:
P
increases in time because
v
increase and because
a
increases.
6.89.
IDENTIFY
and
SET UP:
Energy is
av
.
The total energy expended in one day is the sum of the energy expended in
each type of activity.
EXECUTE:
4
1 day
8.64 10 s
Let
walk
t
be the time she spends walking and
other
t
be the time she spends in other activities;
4
other
walk
.
−
The energy expended in each activity is the power output times the time, so
7
walk
other
(280 W)
(100 W)
1.1 10 J
EP
t
t
t
+
47
walk
walk
(280 W)
(100 W)(8.64 10 s
) 1.1 10 J
+×
−
=
×
6
walk
(180 W)
2.36 10 J
t
4
walk
1.31 10 s
218 min
3.6 h.
t
=
=
EVALUATE:
Her average power for one day is
7
(1.1 10 J)/([24][3600 s]) 127 W.
×=
This is much closer to her
100 W rate than to her 280 W rate, so most of her day is spent at the 100 W rate.
6.90.
IDENTIFY
and
SET UP:
WP
t
=
EXECUTE:
(a)
The hummingbird produces energy at a rate of 0.7 J/s to 1.75 J/s. At 10 beats/s, the bird must
expend between 0.07 J/beat and 0.175 J/beat.
(b)
The steady output of the athlete is (500 W)/(70 kg)
7 W/kg,
=
which is below the 10 W/kg necessary to stay aloft.
Though the athlete can expend 1400 W/70 kg
20 W/kg
=
for short periods of time, no humanpowered aircraft could
stay aloft for very long.
EVALUATE:
Movies of early attempts at humanpowered flight bear out our results.
6.91.
IDENTIFY
and
SET UP:
Use Eq.(6.15). The work done on the water by gravity is
mgh
, where
170 m.
h
=
Solve for
the mass
m
of water for 1.00 s and then calculate the volume of water that has this mass.
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Chapter 6
EXECUTE:
The power output is
9
av
2000 MW
2.00 10 W.
P
==
×
av
W
P
t
Δ
=
Δ
and 92% of the work done on the water
by gravity is converted to electrical power output, so in 1.00 s the amount of work done on the water by gravity is
9
9
av
(2.00 10 W)(1.00 s)
2.174 10 J
0.92
0.92
Pt
W
Δ×
=×
,
Wm
g
h
=
so the mass of water flowing over the dam in 1.00 s must be
9
6
2
1.30 10 kg
(9.80 m/s )(170 m)
W
m
gh
×
=
×
density
m
V
=
so
6
33
1.30 10 m .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Acceleration, Energy, Force, Kinetic Energy, Work

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