Velocity and Loss in Energy onedimensional

[Nuingaer]

Homework Statement

If a 15kg object is released from a height of 11.58m and the objects loss in energy is 302 J, determine its velocity as it enters the container (object is being dropped straight down into a container)
Assume g = 9.8m/s^2
m(object) = 15kg = w(object) = 147N
s = 11.58m
ΔE = -320 J

Homework Equations

Had no idea. But the ones i know that i thought could apply:
ΔKE = KE(final) - KE(initial)
KE = 1/2 * mv^2
GPE = mgh
v^2 = u^2 + 2as

The Attempt at a Solution

Okay, I had no idea how to do this, but i tried to work out as many things as i can then mash em together to get an answer :p

GPE = mgh
=15*9.8*11.58
=1702.26

Then i thought; well if the loss in energy is 320 J then i could take that from the original GPE
1702.26 - 320 = 1382.26 J

and now i could work out the new height by subbing it back in
1382.26 = 15*9.8*h
h=9.403m

And here is where it gets iffy... they say the velocity when it enters the container, but the container isn't at 9.403m its at 0...right? that's what i got from the question anyway. Therefore how did they only lose 320N? So I assumed that I could say its KINETIC energy is the 1382.26 J, and the height of the container is the difference because they say when it "enters".
Therefore..
KE = 1/2 * m * v^2
1382.26 = 1/2 * 15 * v^2
2764.52 = 15 * v^2
v^2 = 184.30
v = 13.58 m/s

We don't have the answer to this question yet, but *a lot* of other people got 13.66m/s as the answer... so since i had no idea how to do this question and no basis to support whether I am right or not, could someone help me out and either point out where i went wrong or if I'm right? thank you!

 

[alphali]

I think they mean by lose of energy that it's due to friction, because when dropping an object from a certain hight the object does't lose energy but the energy is converted from potential to kinetic, and so when the question states that there there was loss in energy i think its due to friction(i might be wrong i don't know). If this is the case then i think you should do this:
Energy (at h=11.58 )-Energy lost=1702.26-320=1382.26
Ek=1/2*mv^2
v=13.57m/s
in this case you take the barrol at hight 0.