fisica1-youn-e-freedman-exercicios-resolvidos

5.85: Denote the magnitude of the acceleration of the block with mass m1
as a; the block
of mass m
2
will descend with acceleration a 2.
If the tension in the rope is T, the
equations of motion are then
m g −
2
T = m a
1
2T
= m2
a
2.
Multiplying the first of these by 2 and adding to eliminate T, and then solving for a gives
m2g
2m2
a =
= g .
2m1
+ m2
2 4m1
+ m2
The acceleration of the block of mass m
2
is half of this, or g m2 ( 4m1
+ m2).
5.86: Denote the common magnitude of the maximum acceleration as a. For block A to
remain at rest with respect to block B, a < µ g s
. The tension in the cord is then
T = ( mA + mB)
a + µ
kg(
mA
+ mB)
= ( mA
+ mB
)( a + µ
kg).
This tension is related to the mass
m by T = mC ( g − a).
Solving for a yields
C
Solving the inequality for
m − µ
k
( mA
+ m
a = g
m + m + m
m
C
yields
)
<
C B
µ s
A B C
g.
m
C
( m
A B
<
+ m )( µ
s
+ µ
k
)
.
1−
µ
s
5.87: See Exercise 5.15 (Atwood’s machine). The 2.00-kg block will accelerate upward
5.00 kg−2.00 kg
at g = 3g
7,
and the 5.00-kg block will accelerate downward at 3g 7.
Let the
5.00 kg+
2.00 kg
initial height above the ground be h
0
; when the large block hits the ground, the small
block will be at a height 2h
0
, and moving upward with a speed given by
2
2
v
0
= 2ah0
= 6gh0
7. The small block will continue to rise a distance v
0
2g
= 3h0
7,
and
so the maximum height reached will be 2h
0
+ 3h0
7 = 17h0
7 = 1.46 m , which is 0.860 m
above its initial height.