kinetic energy

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January 6th 2009, 12:50 PM
Hanga

kinetic energy

Okay so i'll go right to it (been sitting about 2 hours with this question);

An icecube is sliding 3 meters friction free off the ledge of a house roof which has the angle of 27 degrees. What speed will it attain at the edge of the roof?

Just imagine a triangle with the hyp of 3 meters and the angle at the edge is 27degrees. The one and only force that can meddle with the ice cube must be gravity (9.82m/s(square))

So please solve this for me so I can understand what im doing wrong.

The answer is v=square(2*9.82+3*sin(27))

I've got no clue where the 2 comes from...
January 6th 2009, 01:06 PM
Last_Singularity

Quote:

Originally Posted by Hanga

Okay so i'll go right to it (been sitting about 2 hours with this question);

An icecube is sliding 3 meters friction free off the ledge of a house roof which has the angle of 27 degrees. What speed will it attain at the edge of the roof?

Just imagine a triangle with the hyp of 3 meters and the angle at the edge is 27degrees. The one and only force that can meddle with the ice cube must be gravity (9.82m/s(square))

So PLEASE for the LOVE OF GOD solve this for me so I can understand wtf im doing wrong.

The answer is v=square(2*9.82+3*sin(27))

I've got no clue where the 2 comes from...

Are you sure that v=square(2*9.82+3*sin(27))?

Using conservation of energy, the result seems to be:
$KE = PE \rightarrow \frac{1}{2}mv^2 = mgh$

$\frac{1}{2}v^2=gh$

$v = \sqrt{2gh} = \sqrt{2(9.81)(3)(sin27)}$

Alternatively, you can just use kinematics and the answer appears as clear as day:
$v_f^2 = v_0^2 + 2ax$ where $v_f, v_0, a, x$ are the final velocity, initial velocity, acceleration, and distance traveled, respectively.
$v_f^2 = 0 + 2(9.81sin(27))(30) \rightarrow v_f = \sqrt{2(9.81)(3)(sin27)}$
January 6th 2009, 01:18 PM
craig

It looks like you are using the formula $v^2 = u^2 + 2as$ , although I am not sure where you numbers have come from?
January 6th 2009, 01:23 PM
Last_Singularity

Quote:

Originally Posted by craig

It looks like you are using the formula $v^2 = u^2 + 2as$ , although I am not sure where you numbers have come from?

Oops, I meant 3 instead of 30 in
$<br /> v_f^2 = 0 + 2(9.81sin(27))(30) \rightarrow v_f = \sqrt{2(9.81)(3)(sin27)}<br />$

Sorry about that
January 6th 2009, 01:29 PM
craig

Quote:

Originally Posted by Last_Singularity

Oops, I meant 3 instead of 30 in
$<br /> v_f^2 = 0 + 2(9.81sin(27))(30) \rightarrow v_f = \sqrt{2(9.81)(3)(sin27)}<br />$

Sorry about that

Sorry that was aimed at the OP :S

But I'll accept thanks for spotting the error lol ;)
January 6th 2009, 03:07 PM
Hanga

wow dude! I really understand exactly what I was doing wrong here.

This is a very hard question from my book and I acually thought that it would involve simple trigonometry, silly me.

Thanks alot mate! You just made my day!