fisica1-youn-e-freedman-exercicios-resolvidos

2
5.107: a) The forces on the car are the air drag force f
D
= Dv and the rolling friction
force µ mg r
. Take the velocity to be in the + x -direction. The forces are opposite in
direction to the velocity. ∑ F = ma gives
x
x
2
− Dv − µ mg = ma
We can write this equation twice, once with v = 32 m s and
r
2
a = − 0.42 m s and once
with v = 24 m s and a = −0.30 m/s
2 . Solving these two simultaneous equations in the
unknowns D and µ
r
gives µ = 0. r
015 and 0.36
N s 2
m 2
D = ⋅ .
b) n = mg cos β and the component of gravity parallel to the incline is mg sin β,
where β = 2 .2°.
2
For constant speed, mg sin 2.2° − µ
r
mg cos 2.2° − Dv = 0.
Solving for v gives v = 29 m s.
2
c) For angle β , mg sin β − µ mg cos β − Dv 0
r
=
mg(sin
β − µ
r
cos β)
and v =
D
2
The terminal speed for a falling object is derived from Dv − mg 0,
so
v
t
= mg D.
v
v
=
sin β −
t
µ
r
And since
cos β
µ
r
= 0.015,
v vt
=
sin β − (0.015) cos β
t
=
5.108: (a) One way of looking at this is that the apparent weight, which is the same as
the upward force on the person, is the actual weight of the person minus the centripetal
force needed to keep him moving in its circular path:
2
2
mv ⎡
2 (12 m s) ⎤
wapp
= mg − = (70 kg) ⎢(9.8 m s ) −
R
40 m
⎥
⎣
⎦
= 434 N
(b) The cart will lose contact with the surface when its apparent weight is zero; i.e., when
the road no longer has to exert any upward force on it:
2
mv
mg − = 0
R
2
2
2 2
v = Rg = (40 m) (9.8 m s ) = 392 m s
v = 19.8 m s or 20 m s
The answer doesn’t depend on the cart’s mass, because the centripetal force needed to
hold it on the road is proportional to its mass and so is its weight, which provides the
centripetal force in this situation.