fisica1-youn-e-freedman-exercicios-resolvidos

6.47: The power is P = F ⋅v
. F is the weight, mg, so
2
P = (700 kg) (9.8 m s ) (2.5 m s) = 17.15 kW. So, 17 .15 kW 75 kW. = 0.23, or about
23% of the engine power is used in climbing.
6.48: a) The number per minute would be the average power divided by the work (mgh)
required to lift one box,
(0.50 hp) (746 W hp)
= 1.41 s,
2
(30 kg) (9.80 m s ) (0.90 m)
or 84 .6 min. b) Similarly,
(100 W)
(30 kg) (9.80 m s ) (0.90 m)
or
22 .7
min.
2
=
0.378
s,
6.49: The total mass that can be raised is
(40.0 hp) (746 W hp)(16.0 s)
(9.80 m s ) (20.0 m)
2
=
1836 kg
so the maximum number of passengers is = 28.
65.0 kg
2436 kg,
6.50: From any of Equations (6.15), (6.16), (6.18) or (6.19),
Wh ( 3800 N) (2.80 m)
3
P = =
= 2.66×
10 W = 3.57 hp.
t (4.00 s)
( 0.70) P (0.70) (280,000 hp)(746 W hp)
6
6.51: F =
ave =
= 8.1 10 N.
v (65 km h) ((1km h) (3.6 m s))
×
6.52: Here, Eq. (6.19) is the most direct. Gravity is doing negative work, so the rope
must do positive work to lift the skiers. The force F r is gravity, and F = mg,
where
is the number of skiers on the rope. The power is then
P = ( mg)
( v)
cosφ
=
(50) (70 kg) (9.80 m
2
⎛
s ) (12.0 km h) ⎜
⎝
1m s
3.6 km h
⎞
⎟ cos (90.0° −15.0°
)
⎠
4
= 2.96×
10 W.
Note that Eq. (1.18) uses φ as the angle between the force and velocity vectors; in this
case, the force is vertical, but the angle 15 .0°
is measured from the horizontal, so
φ = 90.0° −15.
0°
is used.