fisica1-youn-e-freedman-exercicios-resolvidos

5.40: Differentiating Eq. (5.10) with respect to time gives the acceleration
⎛ k ⎞ −( k m) t −( k m ) t
a = vt⎜
⎟e
= ge ,
⎝ m ⎠
where Eq. (5.9), v t
= mg k has been used.
Integrating Eq. (5.10) with respect to time with y = 0
0 gives
y =
t
∫
0
v [1 − e
t
−(
k m)
t
⎡ ⎛ m ⎞
= vt
⎢t
+ ⎜ ⎟e
⎣ ⎝ k ⎠
⎡ m
= vt
⎢
t −
⎣ k
] dt
−(
k m)
t
−(
k m)
t ⎤
( 1−
e ) .
⎤ ⎛ m ⎞
⎥ − vt⎜
⎟
⎦ ⎝ k ⎠
⎥
⎦
5.41: a) Solving for D in terms of v
t
,
mg (80 kg) (9.80 m s
D = =
2
(42 m s)
2
=
2
v t
)
0.44 kg
2
mg (45 kg)(9.80 m s )
b) v
t
= =
= 42 m s.
D (0.25 kg m)
5.42: At half the terminal speed, the magnitude of the frictional force is one-fourth the
weight. a) If the ball is moving up, the frictional force is down, so the magnitude of the
net force is (5/4)w and the acceleration is (5/4)g, down. b) While moving down, the
frictional force is up, and the magnitude of the net force is (3/4)w and the acceleration is
(3/4)g, down.
m.
5.43: Setting F
net
equal to the maximum tension in Eq. (5.17) and solving for the speed v
gives
Fnet R (600 N)(0.90 m)
v = =
= 26.0 m s,
m (0.80 kg)
or 26 m/s to two figures.
5.44: This is the same situation as Example 5.23. Solving for µ
s
yields
2
2
v (25.0 m s)
µ = = Rg (220 m)(9.80 m s
s
=
2
)
0.290.