Question

If a `5 kg` object moving at `6 m/s` slows to a halt after moving `2 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

The coefficient of kinetic friction is `=0.92`

Explanation:

The mass of the object is `m=5kg`

The initial speed of the object is `u=6ms^-1`

The final speed of the object is `v=0ms^-1`

The distance is `s=2m`

Apply the equation of motion

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

to calculate the acceleration

The acceleration is

`a=(v^2-u^2)/(2s)=(0^2-6^2)/(2*2)=-36/4=-9ms^-2`

According to Newton's Second Law

`F=ma`

The frictional force is `F_r=5*9=45N`

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

The normal force is `N=mg=5*9.8=49N`

The coefficient of kinetic friction is

`mu_k=F_r/N=45/49=0.92`