Question

If a `1/2 kg` object moving at `1/4 m/s` slows to a halt after moving `1/8 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`k~=0,408`

Explanation:

`k : "coefficient of friction"`
`V^2=V_i^2-2*a*x`
`V=0 " if object stops."`
`0=(1/4)^2-cancel(2)*a*1/cancel(8)`
`1/16=a/4`
`a=4 m/s^2 " acceleration"`
`k*N=m*a`
`"N: normal force contacting surface"`
`N=m*g`
`k*cancel(m)*g=cancel(m)*a`
`k=a/g`
`k=4/(9,81)`
`k~=0,408`