fisica1-youn-e-freedman-exercicios-resolvidos

5.82: We take the upward direction as positive. The explorer’s vertical acceleration is
2
− 3.7 m s for the first 20 s. Thus at the end of that time her vertical velocity will be
2
v y
= at = ( −3.7 m s )(20 s) = −74 m s. She will have fallen a distance
⎛ − 74 m s ⎞
d = vavt
= ⎜ ⎟(20 s) = −740 m
⎝ 2 ⎠
and will thus be 1200 − 740 = 460 m above the surface. Her vertical velocity must reach
zero as she touches the ground; therefore, taking the ignition point of the PAPS as
y = 0, 0
2 2
= v + a(
y − y )
v y
0
2
0
2
vy
− v
a =
2( y − y
2
0
0
0 − ( −74 m
=
) − 460
s)
2
= 5.95 m
s
2
or 6.0 m
s
2
which is the vertical acceleration that must be provided by the PAPS. The time it takes to
reach the ground is given by
v − v
t =
a
0
0 − ( −74 m s)
=
2
5.95 m s
= 12.4 s
Using Newton’s Second Law for the vertical direction
F
PAPSv
+ mg = ma
F
PAPSv
= ma − mg = m(
a + g)
= (150kg)(5.95 − ( −3.7))
m
= 1447.5 N or 1400 N
s
2
which is the vertical component of the PAPS force. The vehicle must also be brought to a
stop horizontally in 12.4 seconds; the acceleration needed to do this is
2
v − v0
0 − 33 m s
2
a = =
= 2.66 m s
t 12.4 s
2
and the force needed is F
PAPSh
= ma = (150 kg)(2.66 m s ) = 399 N or 400 N, since there
are no other horizontal forces.