Question

If an object is moving at `8 m/s` over a surface with a kinetic friction coefficient of `u_k=9 /g`, how far will the object continue to move?

Answer 1

The distance is `=3.56m`

Explanation:

The coeffcient of kinetic friction is `mu_k=9/g`

`mu_k=F_r/N`

Let the mass of the object be `=m kg`

Then,

The normal reaction is `N=mg N`

The frictional force is

`F_r=mu_kN=mu_kmg=9/g*mg=9mN`

According to Newton's Second Law

`F=ma`

The acceleration is `a=F/m=F_r/m=-9m/m=-9ms^-2`

The initial velocity is `u=8ms^-1`

Apply the equation of motion

`v^2=u^2+2as`

The final velocity `v=0`

`0=64-2xx9xxs`

The distance is

`s=64/18=3.56m`