Question

During take-off a 8kg model rocket, initially at rest, burns fuel for 6.3s causing its speed to increase from rest to 60m/s during this time despite experiencing a 50N drag.
What is the magnitude of the net force on the rocket?
What is the strength of the thrust?
What is the magntiude of the rocket's acceleration?
What is the rocket's height at the instant it runs out of fuel?

Answer 1

The rocket attains a speed of 60 m/s in a matter of 6.3 s, so it is undergoing an acceleration a of

a = (60 m/s - 0) / (6.3 s) ≈ 9.52 m/s²

During launch, the only forces acting on the rocket are directed vertically:

• its weight (pointing down) with magnitude W,

• drag (pointing down) with magnitude 50 N, and

• thrust (pointing up) with magnitude T.

(a) By Newton's second law, the magnitude of the net force is proportional to the acceleration according to

∑ F = m a

where m is the mass of the rocket, so the magnitude of the net force is

(8 kg) (9.52 m/s²) ≈ 76.2 N

and it is directed upward.

(b) Expanding the equation from part (a) gives

T - W - 50 N = (8 kg) (9.52 m/s²)

The weight is

W = (8 kg) g = (8 kg) (9.80 m/s²) = 78.4 N

so the thrust force is

T ≈ 76.2 N + 78.4 N + 50 N ≈ 205 N

(c) We found this earlier, a ≈ 9.52 m/s².

(d) The rocket's height y at time t is given by

y = 1/2 a t ²

so that after 6.3 s, it will have reached a height of

y = 1/2 (9.52 m/s²) (6.3 s)² = 189 m