fisica1-youn-e-freedman-exercicios-resolvidos

5.88: The floor exerts an upward force n on the box, obtained from n − mg = ma,
or
n = m( a + g). The friction force that needs to be balanced is
2
µ
k
n = µ
km(
a + g)
(0.32)(28.0 kg)(1.90 m s + 9.80 m s ) = 105 N.
2
5.89: The upward friction force must be f = µ n = m g,
and the normal force, which is
s
the only horizontal force on block A, must be n = mAa,
and so a = g . An observer on
the cart would “feel” a backwards force, and would say that a similar force acts on the
block, thereby creating the need for a normal force.
s
A
µ s
5.90: Since the larger block (the trailing block) has the larger coefficient of friction, it
will need to be pulled down the plane; i.e., the larger block will not move faster than the
smaller block, and the blocks will have the same acceleration. For the smaller block,
( 4.00 kg) g (sin30° − (0.25)cos 30°
) − T = (4.00 kg) a,
or 11.11 N − T = (4.00 kg) a,
and
similarly for the larger, 15.44 N + T = (8.00 kg) a,
a) Adding these two relations,
2
26 .55 N = (12.00 kg) a , a = 2.21m s (note that an extra figure was kept in the
intermediate calculation to avoid roundoff error). b) Substitution into either of the above
relations gives T = 2.27 N. Equivalently, dividing the second relation by 2 and
3 15.44 N
subtracting from the first gives T = 11.11 N − ,
2
2
giving the same result. c) The
2
string will be slack. The 4.00-kg block will have a = 2.78 m s and the 8.00-kg block
2
will have a = 1.93 m s , until the 4.00-kg block overtakes the 8.00-kg block and collides
with it.
5.91: a) Let n
B
be the normal force between the plank and the block and n
A
be the
normal force between the block and the incline. Then, n B
= wcosθ
and
nA = nB
+ 3wcosθ
= 4wcosθ.
The net frictional force on the block is
µ
k
( n
A
+ nB
) = µ
k
5wcosθ
. To move at constant speed, this must balance the component
of the block’s weight along the incline, so 3w sinθ
= µ
k5wcosθ,
and
3
3
µ
k
= tanθ
= tan37°
= 0.452.
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