## College Physics (4th Edition)

(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.
(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.