How to get the coefficient of kinetic friction?

[Fatima Hasan]

Homework Statement

Homework Equations

W = Fk d cos θ
Fk = FN μκ

The Attempt at a Solution

Here's my work :
m = 1 kg vi = 10 m/s vf = 0 m/s xi = 0 m xf = 0.1 m d = 0.1 m k=60 N/m μκ=?
The work done be the frictional force :
W = Fκ d cos θ = FN*μκ*d*cosθ = mg μκ d cos 180 = - 0.1*10*1 *μκ = - μκ
- μκ = W = ΔKE + ΔUg + ΔUs
ΔUg = 0 , because the height doesn't change
- μκ = 0.5 m vf^2 - 0.5 m vi^2 + 0.5 k ( xf )^2 - 0.5 k ( xi )^2
- μκ = 0 - 0.5 * 1 * 10^2 + 0.5 * 60 *( 0.1 )^2 - 0
- μκ = - 49.7
μκ = 49.7
I don't know why I've got unreasonable answer . Could somebody help me to solve this problem ?
Thanks for your help

 

[haruspex]

I've got unreasonable answer

It is large, but not impossible. I see no error in your working.

Edit: 60 N/m is a fairly weak spring.

 

[jbriggs444]

When in doubt, sanity-check by another approach.

We have an object moving at 10 m/s coming to a stop over a distance of 0.1 meters. If the deceleration were constant (it's not), that would take 0.02 seconds = 20 milliseconds.

10 m/s in 0.02 seconds is 500 meters/sec² = 50 g's.

The spring, when compressed by 0.1 meters, will produce 6 N. On average (distance-weighted) that would be 3 N. On a one kg mass, that's only 0.3 g.

So a back of the envelope estimate is 49.7 g's from kinetic friction - a coefficient of kinetic friction of 49.7, just as you obtained with the proper analysis.

Edit: [with more significant figures than are warranted for a "g" that is approximated as 10 m/s²].

 

[scottdave]

I also agree with your answer. I think you would benefit from keeping units in your calculations. It helps prevent mistakes, and should give you confidence that you performed the calculations correctly. You may want to read these two Insights:

https://www.physicsforums.com/insights/make-units-work/

https://www.physicsforums.com/insights/10-math-tips-save-time-avoid-mistakes/