Help with work/spring question!

A mass m = 10.0 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5069.0 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.5 m/s.
Work done by spring = 61.25J

2) How far was the spring stretched from its unstretched length?

3) The mass is measured to leave the rough spot with a final speed vf = 1.6 m/s.
How much work is done by friction as the mass crosses the rough spot?

4) What is the length of the rough spot?

5) In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length?

6) In this new scenario, what would the coefficient of friction of the rough patch need to be changed to in order for the block to just barely make it through the rough patch?

2)
v = 3.5 m/s

SE = 1/2*k*x^2
KE = 1/2*m*v^2

Total Energy = SE = KE

1/2*k*x^2 = 1/2*m*v^2
k*x^2 = m*v^2
x^2 = m/k*v^2
x = v*sqrt(m/k)
= 3.5*sqrt(10/5069)
=0.155 m

3)
KE1 = KE2 + frictionWork
FW = KE1 - KE2
= 61.25 - 1/2*m*v2^2
= 61.25 - 1/2*10*1.6^2
= 61.25 - 12.8
= 48.45 J

4)
F = uk*N = uk*m*g
W = F*d = (uk*m*g) * d
d = W/F = W/(uk*m*g)
= 48.45/(0.49*10*9.8)
= 1.0089 m

5)
d = 0.50445m

W = uk*m*g*d
W = 0.49*10*9.8*0.50445
= 24.2237 J

SE = W
1/2*k*x^2 = W
x = sqrt(2*W/k)
= sqrt(2*24.2237/5069)
= 0.0977 m

6)
W = d*uk*m*g

let d = half of rough patch


change uk so that mass barely travels through rough patch instead of halfway
W =d*uk*m*g = (2d)*(x*uk)*m*g
x = 1/2
so reduce uk by half
new uk = 0.245