Question

If an object is moving at `4 ms^-1` 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 `=2.67m`

Explanation:

Let the mass of the object be `=mkg`

The acceleration due to gravity is `g=9.8ms^-2`

The initial velocity is `u=4ms^-1`

The final velocity is `v=0ms^-1`

The coefficient of kinetic friction is `mu_k=3/g`

The normal force is `N=mgN`

The frictional force is `=F_r N`

`mu_k=F_r/N`

`F_r=mu_k*N=3/g*mg=3mN`

According to Newton's Second Law

`F=ma`

`F_r=-ma`

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

Apply the equation of motion,

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

The distance is

`s=(v^2-u^2)/(2a)=(0-4^2)/(2*(-3))=16/6=8/3=2.67m`