Question

If a `1 kg` object moving at `25 m/s` slows down to a halt after moving `625 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`u_k=0,038`

Explanation:

`v_f^2=v_i^2-2*a*x`
`(v/2)^2=v^2-2*a*625`
`v^2/4=v^2-2*a*625`
`v^2=4v^2-8*a*625`
`3v^2=8*a*625`
`v=25 m/s`
`3*25^2=8.a.625`
`3*cancel(625)=8*a*cancel(625)`
`a=3/8 m/s^2" acceleration of the object"`
`F_f=u_k*N" the friction force "`
`N=m*g" N:normal force acting contacting surfaces"`
`F_f=m*a" Newton's the second law"`
`u_k*cancel(m)*g=cancel(m)*3/8`
`u_k=3/(8*9,81)`
`u_k=0,038`