fisica1-youn-e-freedman-exercicios-resolvidos

5.32:
2 2 2 1 2 2
a v0
− v v0
− v
4 0 3 v0
µ
r
= = = = ,
g 2Lg
2Lg
8 Lg
where L is the distance covered before the wheel’s speed is reduced to half its original
3 (3.50 m s)
speed. Low pressure, L = 18.1m;
= 0.0259.
High pressure,
L = 92.9 m;
3
8
2
(3.50 m s)
2
(92.9m)(9.80m s )
= 0.00505.
8
2
(18.1m)(9.80 m s )
2
5.33: Without the dolly: n = mg and F − µ n 0 ( a = 0 since speed is constant).
k
=
F 160 N
m = =
= 34.74 kg
2
µ
kg
(0.47) (9.80 m s )
With the dolly: the total mass is 34 .7 kg + 5.3 kg = 40.04 kg and friction now is rolling
friction, f
r
= µ
rmg.
F − µ mg = ma
r
F − µ
rmg
a = = 3.82 m
m
x
s
2
5.34: Since the speed is constant and we are neglecting air resistance, we can ignore the
2.4 m/s, and Fnet
in the horizontal direction must be zero. Therefore f
r
= µ
rn
=
= F = 200 N
horiz
before the weight and pressure changes are made. After the changes,
( 0.81) (1.42n)
= Fhoriz,
because the speed is still constant and F
net
= 0 . We can simply
divide the two equations:
(0.81µ
r)(1.42n)
Fhoriz
=
200 N.
µ n
(0.81) (1.42)
r
(200 N) = F
horiz
= 230 N