Final velocity and frictional force?

[emmyology]

Homework Statement

a) A jet fighter has two engines each generating 70,000 N of thrust. What is the acceleration of the 15,000 kg jet?


b) The above jet reaches a final velocity of 240 m/s. What is the frictional force on the jet?

Homework Equations

Ff= Fn(mew?)
F=MA
?

The Attempt at a Solution

Okay, so for the first part (a) I got 4.66 m/s squared as my acceleration. I need to know how to get the answer for part b.

 

[danielakkerma]

Hi there, welcome to the forum.
Since the question deals with a flying object, I assume the frictional force mentioned is one of the viscuous property that requires the solution of a differential equation.
If you care for it, and your level of studies satisfies it, it goes something likes this:
A frictional force, acting in a medium such as air, is typically proportional to the velocity, like -k*v, so we get:
$\large F-kv(t) = mv'(t)$
Which leads to the general solution:(assuming the jet started off at zero initial speed)
$\large v(t)=\frac{F}{k}(1-e^{-\frac{kt}{m}})$
The final velocity is obtained somewhere in infinity, or V(final) = F/k;
You're given V_final, so you can easily find the constant k, which will lead you to the frictional force, -k*v, your mu.
Hope that helps,
Daniel

 

[LawrenceC]

But the jet has two engines...part a.

 

[LawrenceC]

I don't know what level class you are taking but the answer may be very simple for part b. The jet ceases to accelerate when the resistance due to air friction equals the thrust of the TWO engines.

 

[danielakkerma]

But the jet has two engines...part a.

I think the question, as posited by the author, implied that:

A jet fighter has two engines each generating 70,000 N of thrust

cumulatively, in unison, all in all, 7*10^4 N;
As for the friction, I doubt the resistive force is constant, in such a query as this.
Daniel

 

[I like Serena]

Welcome to PF, emmyology!

I tend to agree with LawrenceC.
For the answer to (a) I think you need to consider that there are two engines...
And for the answer to (b) I'm also interpreting the question as meaning the friction when the final velocity is reached, at which time the force of friction exactly cancels the force of the two engines.

@danielakkerma: I believe friction with air is typically modeled as being proportional to the square of the velocity.

 

[danielakkerma]

Thanks for clearing that up, I hope I didn't convolute things...
As for the resistance, my initial thoughts leaned towards Stoke's law, whose drag is proportional to the velocity itself, given that the speeds here are not quite so massive. But I could be wrong of course, and I am glad you can point us in the proper direction; I wish we could hear from the OP and the respective elucidations which might entail.
Daniel