Question

If an object is moving at `16` `ms^-1` over a surface with a kinetic friction coefficient of `u_k=80/g`, how far will the object continue to move?

Answer 1

The distance taken for the object to stop will be `1.6` `m`.

Explanation:

The frictional force will be given by:

`F_"frict"=mu_kF_"norm"`

The normal force `F_"norm"=mg`, so:

`F_"frict"=mu_kmg=80/g*mg=80m`

The acceleration of the object will be given by `a=F_"frict"/m=(80m)/m = 80` `ms^-2`. Given that it is a deceleration (acceleration in the opposite direction to the velocity) give it a minus sign: `a=-80` `ms^-2`.

The initial velocity, `u=16` `ms^-1` and the final velocity when the object stops will be `v=0` `ms^-1`.

So `v^2=u^2+2ad`

Rearranging:

`d=-(u^2)/(2d)=-(16^2)/(2*-80)=1.6` `m`