fisica1-youn-e-freedman-exercicios-resolvidos

5.80: If the block were to remain at rest relative to the truck, the friction force would
2
need to cause an acceleration of 2.20 m s ; however, the maximum acceleration possible
2
2
due to static friction is (0.19)(9.80 m s ) = 1.86 m s , and so the block will move relative
2
2
to the truck; the acceleration of the box would be µ
k
g = (0.15)(9.80 m s ) = 1.47 m s .
The difference between the distance the truck moves and the distance the box moves (i.e.,
the distance the box moves relative to the truck) will be 1.80 m after a time
t =
a
truck
2∆x
− a
box
=
(2.20 m
2(1.80 m)
s
2
−1.47 m s
2
)
= 2.22 s.
In this time, the truck moves 1 2 1
2
2
2
a
truckt
=
2
(2.20 m s ) (2.221s) = 5.43 m. Note that an
extra figure was kept in the intermediate calculation to avoid roundoff error.
5.81: The friction force on block A is µ
k
w A
= (0.30)(1.40 N) = 0.420 N, as in Problem 5-
68. This is the magnitude of the friction force that block A exerts on block B, as well as
the tension in the string. The force F must then have magnitude
F = µ
k
( wB
+ wA)
+ µ
kwA
+ T = µ
k
( wB
+ 3wA)
= (0.30)(4.20 N + 3(1.40 N)) = 2.52 N.
Note that the normal force exerted on block B by the table is the sum of the weights of
the blocks.