Question

If a `1 kg` object moving at `7 m/s` slows down to a halt after moving `140 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

I found approx `0.02`

Explanation:

We can start by using:
`color(red)(v_f^2=v_i^2+2ad)`
to get:
`0=7^2+2a*140`
so that:
`a=-49/280=-0.175m/s^2`
We then use Newton´s Second Law `SigmavecF=mveca` along the `x` direction only and with only kinetic friction `f_k` acting to stop the object:
`-f_k=ma`
`-mu_k*N=ma`
where the normal reaction is `N="weight"=mg`
so:
`-mu_k*1*9.8=1*(-0.175)`
so:
`mu_k=0.175/9.8=0.0178~~0.02`