Question

If an object is moving at `12 m/s` over a surface with a kinetic friction coefficient of `u_k=2 /g`, how far will the object continue to move?

Answer 1

`Delta x=36 m`

Explanation:

`"Changing of energy of the kinetic is equal work doing"`
`" by the Friction Force"`
`Delta E_k=W`
`E_(k1)=1/2*m*v^2`
`E_(k2)=0" object stops."`
`Delta E_k=E_(k2)-E_(k1)" "Delta E_k=1/2*m*v^2`
`W_f=F_f*Delta x" work doing by the friction force"`
`W_f=u_k*N*Delta x`
`W_f=Delta E_k`
`u_k*N*Delta x=1/2*m*v^2" "(1)`
`N:"Normal Force acting contacting surfaces. "N=m*g`
`"rearranging the equation (1)"`
`u_k*cancel(m)*g*Delta x=1/2*cancel(m)*v^2`
`u_k=2/g" "v=12 m/s`
`2/cancel(g)*cancel(g)*Delta x=1/2*12^2`
`Delta x=144/4`
`Delta x=36 m`