Question

If an object is moving at `5 m/s` over a surface with a kinetic friction coefficient of `u_k=3 /g`, how far will the object continue to move?

Answer 1

The distance is `=4.17m`

Explanation:

The coeffcient of kinetic friction is `mu_k=3/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=3/g*mg=3mN`

According to Newton's Second Law

`F=ma`

The acceleration is `a=F/m=F_r/m=-3m/m=-3ms^-2`

The initial velocity is `u=5ms^-1`

Apply the equation of motion

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

The final velocity `v=0`

`0=25-2xx3xxs`

The distance is

`s=25/6=4.17m`