energy

[akatsafa]

An airline executive decides to economize by reducing the energy, and thus the amount of fuel, required forlong distance flights. He orders the ground crew to remove the paint from the outer surface of each plane. The paint removed from a single plane has mass 100 kg. (a) If the airplane cruises at an elevation of 29000m, how much energy is saved in not having to lift the paint to that altitude? (b) How much energy is saved per airplane by not having to move the amount of paint from rest to a cruising speed of 230m/s?

I tried these and got 56.84MJ and 31.07MJ, but these are not right. I used the equation E=K+U. I solved for K and then for U, but it's not right. Any suggestions?

Thanks.

[AKG]

An airline executive decides to economize by reducing the energy, and thus the amount of fuel, required forlong distance flights. He orders the ground crew to remove the paint from the outer surface of each plane. The paint removed from a single plane has mass 100 kg. (a) If the airplane cruises at an elevation of 29000m, how much energy is saved in not having to lift the paint to that altitude? (b) How much energy is saved per airplane by not having to move the amount of paint from rest to a cruising speed of 230m/s?

I tried these and got 56.84MJ and 31.07MJ, but these are not right. I used the equation E=K+U. I solved for K and then for U, but it's not right. Any suggestions?

Thanks.I don't know why you'd use that equation. In lifting something, you give it gravitational potential energy. To give it energy, you have to use energy. The energy you use is the energy it gains. Now, the energy you save by not having to lift the paint is the energy that 100kg of paint would have gained if you had lifted it. Calculate the gravitational potential energy increase for 100kg of paint being raised that distance.

\Delta E_{grav} = mg\Delta h = (100kg)(9.81 J/kg\cdot m)(29000m) = 28.4 MJ

For part (b), you use a similar idea, except this time it is the translational kinetic energy which changes. Paint at rest has zero kinetic energy. Paint moving has some. So for it to have that energy, you've exerted energy to speed it up, so if you remove the paint, the energy saved is the energy it would have needed to speed up to that speed.

\Delta E_{k} = E'_{k} - E_{k}

Where E'_{k} is the final kinetic energy (after speeding up) and E_{k} is the initial kinetic energy (at rest).

\Delta E_{k} = \frac{1}{2}mv^2 - 0

You can figure that out.

[akatsafa]

Thank you. I had those equations written down too, but I thought it was a harder question than it really was. Thanks again.