physics question?

Question

I need yet some more help with my physics. I'm looking to understand it, not just for an answer. Thanks in advance to anyone who can help explain how to solve this...

a UFO weighing 1.14*(10^4)kg is about to touch down on the moon, where the acceleration due to gravity is 1.6 m/s^2. At 165m above the surface, the downward velocity (it's moving straight down) is 18 m/s. what is the magnitude of thrust needed to reduce the velocity to 0 the instant the UFO touches down?

So far I'm pretty lost, except that I get that the 1.6m/s is going to be replacing the 9.8 standard here on Earth.

Thanks to anyone who can help me understand this.

Answer 1

If you're lost, it might help to understand that velocity is m/s. Acceleration is m/s^2. In this case it means that a free-falling object on the moon increases its speed by 1.6 m/s each second. So if there's no thrust, the UFO increases from 18 m/s to 19.6 m/s in one second, then to 21.2 m/s the 2nd second, etc.

The weight of the UFO is
W = m*gmoon = 1.14*10^4 kg * 1.6 m/s^2
W = 1.824*10^4 Newtons

Just to offset the acceleration due to gravity, thrust of 1.824*10^4 N is required. With that amount of thrust, the UFO would fall at constant speed. To start to slow it, more thrust is needed to give it a net force upwards. To calculate the required net force, calculate the required deceleration.

The acceleration required is given by
Vf^2 = Vo^2 + 2*a*h
where Vf = 0, Vo = 18 m/s, h = 165 m
(The resulting a should be negative meaning deceleration.)

To achieve this deceleration, the required net force upwards is given by Newton's 2nd:
F = m*a

The thrust needed is F from above plus the thrust required to balance the weight of the UFO -- thrust = F+W.