Question

If an object is moving at `12 m/s` over a surface with a kinetic friction coefficient of `u_k=24 /g`, how far will the object continue to move?

Answer 1

`x=3 meters`

Explanation:

`v^2=v_i^2-2*a*x`
`v=0 " when object is stops"`
`v_i^²=2*a*x`
`k*m*g=m*a`
`24/cancel(g)*cancel(m)*cancel(g)=cancel(m)*a`
`a=24 m/sec^2`
`12^2=2*24*x`
`144=2*24*x`
`144=48.x`
`x=144/48`
`x=3 meters`