fisica1-youn-e-freedman-exercicios-resolvidos

5.67: Consider the forces on the person:
y − y
v
0
2
y
= 3.0 m, a
= v
2
0 y
+ 2a
y
y
( y − y
∑ F
y
= ma
n − mg = ma
= 5.88 m s
0
y
n = 1 .6mg
so a = 0.60 g = 5.88 m
2
,
v
) gives v
0 y
y
= 0, v
y
= ?
= 5.0 m s
2
s
5.68: (a) Choosing upslope as the positive direction:
F
net
= −mg
sin37° − fk
= −mg
sin37° − µ
kmg
cos37°
= ma
and
2
2
a = −( 9.8 m s )(0.602 + (0.30)(0.799)) = −8.25m s
2 2
Since we know the length of the slope, we can use v = v0
+ 2a(
x − x0)
with x = 0
0 and
v = 0 at the top.
2
2
2 2
v = −2ax
= −2(
−8.25 m s )(8.0 m) = 132 m s
0
2 2
v0
= 132 m s = 11.5 m s or 11m s
(b) For the trip back down the slope, gravity and the friction force operate in opposite
directions:
F = −mg
sin37° + µ mg cos 37° = ma
net
k
2
2
a = g(
−sin37° + 0.30 cos37°
) = (9.8 m s )(( −0.602)
+ (0.30)(0.799)) = −3.55 m s
Now
v = 0, x = −8.0 m, x = 0, and
v
0
2
= v
2
0
0
2
+ 2a(
x − x ) = 0 + 2( −3.55 m s )( −8.0 m)
= 56.8 m
v =
2
56.8 m
s
2
2
s
2
0
= 7.54 m s or 7.5 m s