Question

If an object with a mass of `10 kg` is moving on a surface at `45 m/s` and slows to a halt after `4 s`, what is the friction coefficient of the surface?

Answer 1

The coefficient of friction is `=1.15`

Explanation:

Calculate the acceleration with the equation

`v=u+at`

The initial velocity is `u=45ms^-1`

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

The time is `t=4s`

The acceleration is

`a=(v-u)/t=(0-45)/4=-11.25ms^-2`

The mass of the object is `m=10kg`

Apply Newton's Second Law to calculate the frictional force

`F_r=m*|a|=10*11.25=112.5N`

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

The normal force (reaction) is `N=mg=10*9.8=98N`

The coefficient of friction is

`mu_k=F_r/N=112.5/98=1.15`