Work-Energy: Find the angle of the inclined plane

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epr2008
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Homework Statement


A block with mass 5 kg slides down and inclined plane with height 1.5 m and coefficient of kinetic friction .35. The final speed of the block is 4.2 m/s. Find the angle of the inclined plane.

Homework Equations


[tex]{W^{{\rm{net}}}} = {W^{{\rm{grav}}}} - {W^{{\rm{fric}}}} = {K_2} - {K_1}[/tex]

The Attempt at a Solution


This has taken me forever and i still don't think it's right but here it goes.

I have

[tex]{W^{{\rm{net}}}} = mg(\Delta y) - {\mu _k}mgd\cos (\varphi ) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]

Then since initial velocity is 0

[tex]{W^{{\rm{net}}}} = mg(\Delta y) - {\mu _k}mgd\cos (\varphi ) = \frac{1}{2}mv_2^2[/tex]

I'm sure there is another way to find d but i couldn't think of any besides the law of sines.

[tex]\frac{d}{{\sin (90)}} = d = \frac{{\Delta y}}{{\sin (\varphi )}}[/tex]

So now I have

[tex]{W^{{\rm{net}}}} = mg(\Delta y) - {\mu _k}mg(\Delta y)\cot (\varphi ) = \frac{1}{2}mv_2^2[/tex]

And solving

[tex]\cot (\varphi ) = \frac{{mg(\Delta y) - \frac{1}{2}mv_2^2}}{{{\mu _k}mg(\Delta y)}}[/tex]

I am using the calculator on my phone so I don't know if it is right but

[tex]\varphi = 41.186^\circ[/tex]

Would anyone care to help me out and tell me if I'm at least in the ballpark?
 
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