Question

If a `2 kg` object moving at `1/2 m/s` slows to a halt after moving `5 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`mu_k = 0.002548`

Explanation:

Well, the force of friction `F_f` is defined as the following...

`F_f = mu_k N`

And the normal is given by...

`N = mg`

`N = (2)(9.81)`

`N = 19.62N`

So now, we need to calculate an acceleration to get the force of friction. We are assuming in this question that `F_(drive) = F_f` because all the energy that the mass possess is lost to friction.

`v^2=u^2-2as`

`0^2 = 0.5^2 -2(5)a`

`0 = 0.25-10a`

`a = (-0.25)/(10)`

`a= -0.025 ms^-2`

`F_f=ma`

`F_f=(1)abs(-0.025)`

`F_f=0.025N`

`mu_k = F_f/N`

`mu_k = 0.025/9.81`

`mu_k = 0.002548`