Question

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

Answer 1

The distance is `=0.5m`

Explanation:

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

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

The distance is `s=?m`

The coefficient of friction is

`mu_k=F_r/N`

Let the mass be `=mkg`

Then

the normal reaction is `N=mgN`

So,

`F_r=mu_k*mgN`

According to Newton's Second Law

`F=ma`

Therefore,

`ma=-mu_kmg`

The acceleration is `a=-mu_kg` , the negative sign indicates a deceleration

We apply the equation of motion

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

`0=1^2-2*1/g*g*s`

`s=1/2=0.5m`