#### Answer

(a) $\sum F = 1.52~N$
(b) $F = 2.37~N$
(c) $\sum F = 0.85~N$
(d) At the top of the trajectory, the net force on the rocket was the gravitational force of $0.85~N$. The acceleration was $g$, that is, $9.80~m/s^2$ downward.

#### Work Step by Step

(a) We can find the net force:
$\sum F = ma$
$\sum F = (0.087~kg)(17.5~m/s^2)$
$\sum F = 1.52~N$
(b) We can find the upward force $F$ on the rocket by the burning fuel:
$F-mg = \sum F$
$F = \sum F+mg$
$F = 1.52~N+(0.087~kg)(9.80~m/s^2)$
$F = 2.37~N$
(c) After the fuel was spent, the net force on the rocket was the gravitational force:
$\sum F = mg$
$\sum F = (0.087~kg)(9.80~m/s^2)$
$\sum F = 0.85~N$
(d) At the top of the trajectory, the net force on the rocket was the gravitational force of $0.85~N$. The acceleration was $g$, that is, $9.80~m/s^2$ downward.