fisica1-youn-e-freedman-exercicios-resolvidos

mg mg
5.103: Without buoyancy, kv
t
= mg,
so k = = .
vt
0.36 s
With buoyancy included there is the additional upward buoyancy force B, so
B kv = mg
− t
⎛ 0.24 m s ⎞
B = mg − kvt
= mg⎜1
− ⎟ = mg 3
⎝ 0.36 m s ⎠
5.104: Recognizing the geometry of a 3-4-5 right triangle simplifies the calculation. For
instance, the radius of the circle of the mass’ motion is 0.75 m.
4 4
a) Balancing the vertical force, T − T = w,
so
U 5
L 5
5
T
5 80.0 N (4.00 kg) (9.80 m s
2
L
= TU
− w = −
) = 31.0 N.
4
4
2
3 3
4π
R
b) The net inward force is F = T + T = 66.6 N. Solving F = ma = m
2 for
5
U
5
L
the period T,
mR (4.00 kg) (0.75 m)
T = 2 π = 2π
= 1.334 s,
F
(66.6 N)
or 0.02223 min, so the system makes 45.0 rev/min. c) When the lower string becomes
slack, the system is the same as the conical pendulum considered in Example 5.22. With
2
cos β = 0.800, the period is T = 2π
(1.25 m) (0.800) (9.80 m s ) = 2.007 s, which is the
same as 29 .9 rev min. d) The system will still be the same as a conical pendulum, but
the block will drop to a smaller angle.
rad
T