# How to get the coefficient of kinetic friction?

1. Jan 2, 2018

### Fatima Hasan

1. The problem statement, all variables and given/known data

2. Relevant equations
W = Fk d cos θ
Fk = FN μκ
3. 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 ?

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2. Jan 2, 2018

### haruspex

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

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

Last edited: Jan 2, 2018
3. Jan 2, 2018

### 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].

4. Jan 3, 2018