fisica1-youn-e-freedman-exercicios-resolvidos

5.14: a) In level flight, the thrust and drag are horizontal, and the lift and weight are
vertical. At constant speed, the net force is zero, and so F = f and w = L.
b) When the
plane attains the new constant speed, it is again in equilibrium and so the new values of
the thrust and drag, F′ and f ′ , are related by F ′ = f ′; if F ′ = 2F,
f ′ = 2 f . c) In order to
increase the magnitude of the drag force by a factor of 2, the speed must increase by a
factor of 2 .
5.15: a)
The tension is related to the masses and accelerations by
T − m g = m a
T − m2g
= m2a2.
b) For the bricks accelerating upward, let a
1
= −a2
= a (the counterweight will
accelerate down). Then, subtracting the two equations to eliminate the tension gives
( m −m
) g = ( m + m ) a,
or
2
1
1
m2
− m1
2 ⎛ 28.0 kg −15.0 kg ⎞
2
a = g = 9.80 m s ⎜
⎟ = 2.96 m s .
m2
+ m1
⎝ 28.0 kg + 15.0kg ⎠
c) The result of part (b) may be substituted into either of the above expressions to find
the tension T = 191 N. As an alternative, the expressions may be manipulated to
eliminate a algebraically by multiplying the first by m2
and the second by m1
and adding
(with a2 = −a1
) to give
T ( m + m ) − 2m m g = 0, or
2m1m 2g
2(15.0 kg) (28.0 kg) (9.80 m
T = =
m1
+ m2
(15.0 kg + 28.0 kg)
In terms of the weights, the tension is
2m2
2m1
T = w1
= w2
.
m + m m + m
1
2
1
1
1
1
2
1
2
2
1 2
2m 1
m1
+ m2
2
s )
= 191 N.
If, as in this case, m
2
> m1, 2m2
> m1
+ m2
and < , so the tension is greater
than w1
and less than w2;
this must be the case, since the load of bricks rises and the
counterweight drops.