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Work and Kinetic Energy
613
6.49.
IDENTIFY:
av
W
P
t
Δ
=
Δ
. The work you do in lifting mass
m
a height
h
is
mgh
.
SET UP:
1 hp
746 W
=
EXECUTE:
(a)
The number per minute would be the average power divided by the work (
mgh
) required to lift one
box,
2
(0.50 hp) (746 W hp)
1.41 s,
(30 kg) (9.80 m s ) (0.90 m)
=
or
84.6 min.
(b)
Similarly,
2
(100 W)
0.378 s,
(30 kg) (9.80 m s ) (0.90 m)
=
or
22.7 min.
EVALUATE:
A 30kg crate weighs about 66 lbs. It is not possible for a person to perform work at this rate.
6.50.
IDENTIFY
and
SET UP:
Use Eq.(6.15) to relate the power provided and the amount of work done against gravity in
16.0 s. The work done against gravity depends on the total weight which depends on the number of passengers.
EXECUTE:
Find the total mass that can be lifted:
av
,
Wm
g
h
P
tt
Δ
==
Δ
so
av
Pt
m
gh
=
4
av
746 W
(40 hp)
2.984 10 W
1 hp
P
⎛⎞
×
⎜⎟
⎝⎠
4
3
av
2
(2.984 10 W)(16.0 s)
2.436 10 kg
(9.80 m/s )(20.0 m)
m
gh
×
=
×
This is the total mass of elevator plus passengers. The mass of the passengers is
33
2.436 10 kg 600 kg 1.836 10 kg.
×−
=
×
The number of passengers is
3
1.836 10 kg
28.2.
65.0 kg
×
=
28 passengers can ride.
EVALUATE:
Typical elevator capacities are about half this, in order to have a margin of safety.
6.51.
IDENTIFY:
Calculate the gallons of gasoline consumed and from that the energy consumed. Find the time
t
Δ
for the
trip and use
av
W
P
t
Δ
=
Δ
, where
W
Δ
is the energy consumed.
SET UP:
200 km
124 mi
=
EXECUTE:
(a)
The gallons of gasoline consumed is
124 mi
4.13 gal
30 mi/gal
=
. The energy consumed is
99
(4.13 gal)(1.3 10 J/gal)
5.4 10 J
×=
×
.
(b)
The time for the trip is
124 mi
2.07 h
7450 s
60 mi/h
.
9
5
av
7.2 10 W
720 kW
7450 s
W
P
t
Δ×
=
×
=
Δ
.
EVALUATE:
The rate of energy consumption is
3
720 10 W
970 hp
746 W/hp
×
=
.
6.52.
IDENTIFY:
Apply
PF
v
=
!
.
F
!
is the force
F
of water resistance.
SET UP:
746 W
=
. 1 km/h
0.228 m/s
=
EXECUTE:
6
(0.70)
(0.70) (280,000 hp)(746 W hp)
8.1 10 N.
(65 km h) ((0.228 m/s) (1 km/h))
P
F
v
=
×
EVALUATE:
The power required depends on speed, because of the factor of
v
in
v
=
!
and also because the
resistive force increases with speed.
6.53.
IDENTIFY:
To lift the skiers, the rope must do positive work to counteract the negative work developed by the
component of the gravitational force acting on the total number of skiers,
rope
sin
FN
m
g
α
=
.
SET UP:
rope
vFv
!
EXECUTE:
( )
rope
rope
cos
vN
m
g
v
φ
⎡⎤
+
⎣⎦
.
()
(
)
2
rope
1 m/s
50 riders 70.0 kg 9.80 m/s
cos75.0
12.0 km/h
3.60 km/h
P
⎡
⎤
=
⎢
⎥
⎣
⎦
”
.
4
rope
2.96 10 W
29.6 kW
P
=×
=
.
EVALUATE:
Some additional power would be needed to give the riders kinetic energy as they are accelerated from
rest.
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Chapter 6
6.54.
IDENTIFY:
Relate power, work and time.
SET UP:
Work done in each stroke is
WF
s
=
and
av
PW
t
=
.
EXECUTE:
100 strokes per second means
av
100
PF
s
t
=
with
1.00 s,
2
tF
m
g
=
=
and
0.010 m.
s
=
av
0.20 W.
P
=
EVALUATE:
For a 70 kg person to apply a force of twice his weight through a distance of 0.50 m for 100 times per
second, the average power output would be
5
7.0 10 W
×
. This power output is very far beyond the capability of a
person.
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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, Energy, Kinetic Energy, Mass, Power, Work

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