Potential energy and conservation

[lo31415926535]

Homework Statement

A 10.0 kg block is released from point A in Figure P8.57. The track is frictionless except for the portion BC, with a length of 6.00 m. The block travels down the track, hits a spring of force constant k = 2250 N/m, and compresses it 0.300 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between surface BC and the block.

image can be found here: http://www.webassign.net/sb5/p8-57.gif

Homework Equations

Ug=mgh
Uel=(1/2)kx^2
K=(1/2)mv^2
w=FΔx
Fk=μmg

The Attempt at a Solution

initial potential energy - work done by friction = final elastic energy
Ug - Work by friction = Uel
mgh - FΔx = (1/2)kx^2
mgh - μmgΔx = (1/2)kx^2
μ = [mgh - (1/2)kx^2]/(mgΔx)
plugging in the numbers…
μ = [(10 kg)(9.8 m/s^2)(3 m) - .5(2025 N/m)(.3 m)^2]/[(10 kg)(9.8 m/s^2)(6 m)]
and I got μ= 0.345, but this is incorrect
and I am not sure why.
Can someone please help point out my errors?
Thanks!

 

[Doc Al]

μ = [(10 kg)(9.8 m/s^2)(3 m) - .5(2025 N/m)(.3 m)^2]/[(10 kg)(9.8 m/s^2)(6 m)]

This might just be a typo, but you have a different value for k here than earlier.

 

[lo31415926535]

This might just be a typo, but you have a different value for k here than earlier.

Oops that was the problem! I typed it in wrong when i was making my calculations.
Thanks so much