Solve Physics 12 Problem: Helicopter of 5500 kg Reaches 5500 m in 900s

[seiferseph]

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!

 

[dextercioby]

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.

 

[seiferseph]

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]

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

 

[seiferseph]

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]

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

Daniel.

 

[Doc Al]

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.)

 

[seiferseph]

great, thanks!