Question

If an object with a mass of `10 kg` is moving on a surface at `18 m/s` and slows to a halt after `4 s`, what is the friction coefficient of the surface?

Answer 1

`u_k~=0,459`

Explanation:

`v_l=v_i-a*t`
`v/2=v-a*t" "v-v/2=a*t" "v/2=a*t`
`18=a*4" "a=9/2 m/s^2`
`F=m*a" "u_k*N=m*a" "u_k*cancel(m)*g=cancel(m)*a`
`u_k=a/g" "u_k=9/(2*9,81) " "u_k~=0,459`