Question

If a `4 kg` object moving at `6 m/s` slows to a halt after moving `35 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`\mu_k=0.0525`

Explanation:

Since the object is moving on a plane surface, the kinetic friction experienced by the object will be `\mu_kmg` implying that the de-acceleration due to friction will be `\mu_kg`

So, using the third kinematic equation
`v^2=v_o^2+2ax`

The object is said to have come to a halt so that means `v=0ms^-1`
So, `v_o^2=2ax` (we ignored the negative sign since we know why that's there).

So, we know it travels a distance of `35m` before it stops so `x=35m`. We know it initially had a speed of `6ms^-1` so `v_o=6ms^-1`.

So, `a=v_o^2/(2x)=6^2/(2*35)=36/70=0.51`
So, `a=0.51`. But `a=\mu_kg` So, `0.51=\mu_k*9.8\implies\mu_k=0.51/9.8`

Solving it yourself, you now understand that things look odd even if they are right.