fisica1-youn-e-freedman-exercicios-resolvidos

2
5.16: Use Second Law and kinematics: a = g sin θ,
2ax
= v , solve for θ .
g sin θ = v
2
2x,
or
2
2
2
( v 2gx) = arcsin[(2.5 m s) [(2)(9.8 m s )(1.5 m)]], = 12.3°
.
θ = arcsin
θ
5.17: a)
b) In the absence of friction, the net force on the 4.00-kg block is the tension, and so
2
the acceleration will be (10.0 N) (4.00 kg) = 2.50m s . c) The net upward force on the
suspended block is T − mg = ma,
or m = T ( g + a).
The block is accelerating downward,
2
2
2
so a = −2.50 m s , and so m = (10.0 N) (9.80 m s −2.50 m s ) = 1.37 kg.
d) T = ma + mg,
so T < mg,
because a < 0.
5.18: The maximum net force on the glider combination is
12 ,000 N − 2×
2500 N = 7000 N,
7000 N
2
so the maximum acceleration is a = 5.0m s .
max
=
1400 kg
2
v
a) In terms of the runway length L and takoff speed v a = < a , so
2
v
L >
2 a
max
,
2L
max
2
(40 m s)
=
= 160 m.
2
2(5.0 m s )
b) If the gliders are accelerating at a
max,
from
2
T − F = ma,
T = ma + F = (700 kg)(5.0 m s ) + 2500 N 6000 N. Note that this is
drag drag
=
exactly half of the maximum tension in the towrope between the plane and the first
glider.
5.19: Denote the scale reading as F, and take positive directions to be upward. Then,
w ⎛ F ⎞
F − w = ma = a, or a = g⎜
−1⎟.
g ⎝ w ⎠
2
2
a) a = ( 9.80 m s )((450 N) (550N) − 1) = −1.78 m s , down.
2
2
b) a = (9.80 m s )((670 N) (550 N) −1)
= 2.14 m s , up. c) If F = 0 , a = −g
and the
student, scale, and elevator are in free fall. The student should worry.