- #1

Nuingaer

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## 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!

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