How Much Work Does a Plane's Engine Perform to Reach 3200m at 65m/s?

[Some_Thing]

Homework Statement

A 1500 kg plane, initially at rest, leaves an airfield and a short time later it is at an elevation of 3200 m traveling at 65 m/s. What is the minimum work done by the plane's engine in this time? (Ignore air resistance)

Homework Equations

W=Fd, F=ma, and most certainly some other ones.

The Attempt at a Solution

This is what I understand from the question:

Vox = 0 m/s

Voy = 0 m/s

dy = 3200 m

m = 1500 kg

I assume that at 3200m, the plane starts to fly only horizontally, so "65 m/s" would be Vfx.


The problem is that this seems insufficient for any of the kinematics formulas, and I don't see how I could use it in different areas (like dynamics).


What am I missing here? Can someone please give me a push in the right direction?

Any help is greatly appreciated,

Thanks.

 

[Cyosis]

Kinematic equations won't do you any good here. Can you tell me which quantity is conserved in this system?

 

[Some_Thing]

Kinematic equations won't do you any good here. Can you tell me which quantity is conserved in this system?

Ah, I see, its one of those conservation of energy questions, isn't it?

Change in Ek + Change in Ep = the correct answer (5.0 x 10^7)

Thanks!


One thing bugs me though. Shouldn't these two add up to 0?

 

[rock.freak667]

Ah, I see, its one of those conservation of energy questions, isn't it?

Change in Ek + Change in Ep = the correct answer (5.0 x 10^7)

Thanks!


One thing bugs me though. Shouldn't these two add up to 0?

Why would they add up to zero?

There is no potential energy on the ground. The change in kinetic energy on the ground is converted to potential energy at 3200m in the air and kinetic energy to make it travel at 65m/s