Calculating Velocity Using Work and Friction

[rsala]

Homework Statement

A skier moving at 5.00 m/s encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. the patch is 2.9 meters long.
what is her velocity after passing the patch?

problem must be solved by using work, avoid Newtons

Homework Equations

work = fs
work = K_{final} - K_{initial} =- \frac{1}{2}m v^{2}_{final}- \frac{1}{2}m v^{2}_{initial}

The Attempt at a Solution

let s be displacement (2.9m)

-\mu_{k}mg*s = - \frac{1}{2}m v^{2}_{final}- \frac{1}{2} m v^{2}_{initial}<br />

m is irrelevant, factor it out and cancel.

<br /> -\mu_{k}g*s = - \frac{1}{2} v^{2}_{final}- \frac{1}{2} v^{2}_{initial}

solve for v_{final}

\frac{-\mu_{k}gs+.5v^{2}_{initial}}{-.5} = v^{2}_{final}<br />

<br /> \sqrt{-2(-\mu_{k}gs+\frac{1}{2}v^{2}_{initial})} = v_{final}<br />

<br /> \sqrt{ -12.4952}<br />
cannot take sqrt of negative number.
i can't go beyond this part, is there a solution?

 

[Dick]

<br /> \mu_{k}g*s = \frac{1}{2} v^{2}_{initial}- \frac{1}{2} v^{2}_{final}

Where are you finding all the extra minus signs?

 

[rsala]

in which part

 

[Dick]

in which part

All over. Look at your first expression for K_(final)-K(initial). That's supposed to be the difference of two kinetic energies, not the negative of their sum.

 

[rsala]

o I've messed my formula up =/
i got correct answer now, thanks