Question

If a `1 kg` object moving at `14 m/s` slows down to a halt after moving `7 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`W_"net" = -3/8(14)^2 J = mu_k (mg)*d ~~ 70mu_k`
Solve for `mu_k ~~ 21/20` This seem too excessive of a coefficient of kinetic friction... See below for possible explanation...

Explanation:

We can use the work energy theorem, which states that the net work done is equal to the change in Kinetic Energy:
`W_"net" = DeltaKE = 1/2 m(v_f^2 - v_i^2) = 1/2 m((1/2v_i)^2 - v_i^2)`
`W_"net" = 1/2 m(-3/4v_i^2)= -3/8mv_i^2`;
Now with `m = 1; v_i = 14 m/s`
`W_"net" = -3/8(14)^2 J`

Now let's draw a Free Body Diagram:
From FBD we see that the horizontal force slowing the object is the force of friction acting in the opposite direction of motion. And knowing that `W = F*d`
`F_f = mu_k (mg)` this the force doing work hence:
`W_"net" = -3/8(14)^2 J = mu_k (mg)*d ~~ 70mu_k`
Solve for `mu_k ~~ 21/20` Now this amount of kinetic friction is
excessive. What went wrong? Well the general approach is correct but the assumption that friction is the only one consuming energy in slowing the object may not be correct. One can argue Heat and Sound will also take energy away lowering the energy available for friction and hence approaching a reasonable answer of `mu_k < .4`