Frictional Forces-finding velocity and distance

[gungo]

Homework Statement

A 0.170 kg hockey puck is initially moving at 21.2 m/s [W] along the ice. The coefficient
of kinetic friction for the puck and the ice is 0.005.

(a) What is the speed of the puck after traveling 58.5 m?
(b) After being played on for a while, the ice becomes rougher and the coefficient of
kinetic friction increases to 0.047. How far will the puck travel if its initial and final
speeds are the same as before?

Homework Equations

v2^2=v1^2+2ad
Fnet=ma
Ff=μkFn

The Attempt at a Solution

a) FN=9.8(0.170)=1.666
Ff=0.005(1.6660)=0.00833
Fnet=ma
0.00833=0.170a =0.049
v2^2=v1^2+2ad
v2^2=(21.2)^2+2(0.049)(58.5)
v2=21.3, but the answer is 21.1

b)Ff=0.047(1.666)
Ff=0.078302
0.078302=0.170a=0.4606
v2^2=v1^2+2ad
I'll use the correct v2
21.1^2=21.2^2+2(0.4606)d
-4.23=0.9212d
d=-4.59 but the answer is 6.24
Not sure where I'm going wrong?

 

[TomHart]

Does friction acting on the puck produce a positive or negative acceleration?

 

[gungo]

Does friction acting on the puck produce a positive or negative acceleration?

Negative, so v2^2=(21.2)^2+2(-0.049)(58.5), v2=21.1
21.1^2=21.2^2+2(-0.4606)d
-4.23=-0.9212d
d=4.59...That helped fix my mistake in (a) but how about b?

 

[TomHart]

For part b, you really should use more significant digits than 21.1. I think a more accurate number was 21.06. That should resolve that problem - I think.

 

[gungo]

For part b, you really should use more significant digits than 21.1. I think a more accurate number was 21.06. That should resolve that problem - I think.

Oh wow that made a difference than you so much!