# Homework Help: Work and Power for a Plane Engine

1. Jan 9, 2012

### jojo711

1. The problem statement, all variables and given/known data
A 2175 kg test plane has an engine that can deliver a maximum power of 5.28×105 watts. The pilot hits the throttle when the plane is flying at 46.8 m/s and maintains maximum power for 6.5 seconds.
What is the plane's initial kinetic energy?
What is the plane's change in kinetic energy?
What is the plane's final kinetic energy?
What is the new speed of the plane after the 6.5 seconds?

3. The attempt at a solution
I got the initial kinetic energy to be 1.5 x 10^6 J, but do not know how to get the other parts. Please help!

2. Jan 9, 2012

### Delphi51

I got a larger answer for the initial KE. Maybe post your work so I can check the details.
The question doesn't mention air resistance, so you must be expected to ignore it. The engine power must be 100% converted to kinetic energy since no efficiency factor is given. So, all you have to do is calculate the amount of work or energy that power amounts to over the 6.5 seconds. Do you have a formula relating power and work?
Add that work energy onto the initial energy, set it equal to Ek = ½mv² and solve for the final v.

3. Jan 9, 2012

### jojo711

could you please explain more because I really don't understand what you are trying to say but the problems I have to work with:

1/2mv^2
w=fd
p=w/t

4. Jan 9, 2012

### Delphi51

So you have KE = 1/2mv^2 = 1/2*2175*46.8² = ?
I don't get 1.5 x 10^6. Better check it.

Solve your P = W/t formula for W so you can calculate the work done by the engine.

5. Jan 9, 2012

### jojo711

oh i actually changed it and got

2.38×10^6 J

and then I found the other two answers, and they were both right I got the plane's kenetic energy was 3.43X10^6 J and the final was 5.81x10^6 J.

I still cannot find the last one. What is the new speed of the plane after the 6.5 seconds?
The speed so I have to find velocity? so maybe PE=1/2mv^2?

6. Jan 9, 2012

### Delphi51

Great!
Yes, use so maybe KE=1/2mv^2.