Question

If an object is moving at `15 m/s` over a surface with a kinetic friction coefficient of `u_k=15 /g`, how far will the object continue to move?

Answer 1

`x=7,5m`

Explanation:

`"you can use the formula "v^2=v_i^2-2*a*x`
`"v=0 ,if object stops"`
`0=v_i^2-2*a*x`
`v_i^2=2*a*x" (1)"`
`"where "v_i=15m/s" ; "a:"acceleration ; x:displacement"`

`F_f=15/g*m*g=15*m" friction force"`
`F=m*a" the Newton's law"`
`F=F_f`
`m*a=m*15`
`a=15m/s^2`

`"now use equation (1)"`
`15^2=2*15*x`
`15=2*x`
`x=7,5m`