fisica1-youn-e-freedman-exercicios-resolvidos

5.96: (a) 80 mi h is 35 .7 m s in SI units. The centripetal force needed to keep the car on
the road is provided by friction; thus
2
mv
µ mg =
s
r =
r
2
2
v (35.7 m s)
=
= 171m or 170 m
2
µ g (0.76)(9.8 m s )
s
(b) If µ = 0.20
s
:
2
2
2
v = rµ
sg
= (171m) (0.20) (9.8 m s ) = 335.2 m
v = 18.3 m s or about 41mi h
(c) If µ = 0.37
s
:
2
2
2
v = (171m) (0.37) (9.8 m s ) = 620 m
2
s
v = 24.9 m s or about 56 mi h
The speed limit is evidently designed for these conditions.
s
2
5.97: a) The static friction force between the tires and the road must provide the
centripetal acceleration for motion in the circle.
2
v
µ mg s
= m
r
v1
v2
m, g, and r are constant so = , where 1 refers to dry road and 2 to wet
µ µ
road.
1
µ , s 2
=
2
µ
s1
so v2
= (27 m s) 2 = 19 m s
b) Calculate the time it takes you to reach the curve
v = 27 m s, v = 19 m s, x − x = 800 m, t = ?
0x
x
s1
0
s2
⎛ v0x
+ vx
⎞
x − x0
= ⎜ ⎟ t gives t = 34.7 s
⎝ 2 ⎠
During this time the other car will travel x − x0 = v0
x
t = (36 m s) (34.7s) = 1250 m. The
other car will be 50 m behind you as you enter the curve, and will be traveling at nearly
twice your speed, so it is likely it will skid into you.
5.98: The analysis of this problem is the same as that of Example 5.22; solving for v in
2
terms of β and R , v = gR tan β = (9.80 m s ) (50.0) tan 30.0°
= 16.8 m s, about
60 .6 km
h.