Newton's laws and forces

Question

A 90 kg rocket man with a 50 kg jet pack is burning up his fuse up there alone. If the jet pack's thruster puts out 2 kilo-Newtons of force and is pointed 40 degrees below horizontal, what is the magnitude and direction of the man's net acceleration?

Answer 1

This is an example of Newton's second law, where F = m * a. Assuming there are no other forces acting on the rocket man, the acceleration can be found by dividing both sides of the equation by m, so that

a = F / m.

The total mass of the rocket man and the jet pack combined is 90 kg + 50 kg = 140 kg, so the magnitude of the acceleration is

a = (2 kN) / (140 kg) = (2000 N) / (140 kg) = 14.29 m/s^s.

The direction of the acceleration will be in the same direction as the force, 40 degrees below the horizon.