# Physics 12 Problem

I have a quick physics 12 problem

A helicopter of 5500 kg starting from rest reaches a height of 5500 m with a velocity of 210 m/s. If this happens in 900 seconds, what is the Power?

i'm not sure how to do this one, i used P = W/t and solved for work being Ef - Ei. but is there no initial energy? so is it just

P = Ek + Ep (both final)/t
to solve? thanks!

Last edited:

dextercioby
Homework Helper
You can assume that initially,the gravitatostatic potential energy is zero and that the gravity field doesn't vary significantly over those 5.5 Km,so you can set the gravitatostatic PE to "mgh".

Daniel.

dextercioby said:
You can assume that initially,the gravitatostatic potential energy is zero and that the gravity field doesn't vary significantly over those 5.5 Km,so you can set the gravitatostatic PE to "mgh".

Daniel.

so can you solve like that for the power? Power = potential + kinetic / time ?

Doc Al
Mentor
seiferseph said:
i'm not sure how to do this one, i used P = W/t and solved for work being Ef - Ei. but is there no initial energy? so is it just

P = Ek + Ep (both final)/t
to solve? thanks!
That's correct. (Assuming you measure the PE from ground level.)
$P = \Delta E_t /\Delta t$

Doc Al said:
That's correct. (Assuming you measure the PE from ground level.)
$P = \Delta E_t /\Delta t$

so potential is m*g*h with height being 5500, and kinetic is 1/2 m*v^2 with v being 210? thanks.

dextercioby
Homework Helper
Yep.The numbers are not too realistic,but it's okay so far.

Daniel.

Doc Al
Mentor
seiferseph said:
so potential is m*g*h with height being 5500, and kinetic is 1/2 m*v^2 with v being 210? thanks.
That's right. (Realize that you are calculating the power required to raise the helicopter as stated in the given time; the actual power of the engine must be greater than that, since energy is wasted as thermal energy and air movement.)

great, thanks!