Question

If an object is moving at `120 m/s` over a surface with a kinetic friction coefficient of `u_k=7 /g`, how much time will it take for the object to stop moving?

Answer 1

I found `17.1sec`

Explanation:

Kinetic friction `f_k` is responsible for introducing a deceleration given by (Newton's Second law):
`-f_k=ma`
`-u_k*N=ma`
`-u_k*cancel(m)g=cancel(m)a`
So:
`a=-7/g*g=-7`
We can now use:
`color(red)(v_f=v_i+at)`
To get:
`0=120-7t`
`t=120/7=17.1 sec`