Question

If a `3 kg` object moving at `9 m/s` slows down to a halt after moving `27 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`mu=0.15`

Explanation:

The work being done on the object is Work due to friction so the following equation is going to be used:

`color(white)(aaaaaaaaaaaa)`Equation (a) `W_f = DeltaKE`

We can rewrite Equation (a) if we break down both sides step-by-step to become:

`color(white)(aaaaaaaaaaaaaaaaaaa)`Equation (b)
`color(white)(aaaaaa)` `(mu*mg)*d*costheta = (1/2mv_f^2 - 1/2mv_i^2)`

`mu = "coefficient of kinetic friction"`
`m = "mass (kg)"`
`g = "acceleration due to gravity" (m/s^2)`
`d = "displacement"(m)`
`theta = "angle between friction and displacement"`
`v_f = "velocity final"`
`v_i = "velocity initial"`

Since our object stopped, its final velocity becomes `0` and therefore `"final KE"` becomes `0`. Friction and displacement are opposite one another so `cos(180^@) = -1`. Mass cancels on both sides. Rearrange, plug in, and solve.

`-mu*g*d= - 1/2v_i^2`

`mu=(0.5*9^2)/(9.8*27) = 40.5/264.6 = 0.15`

Answer: 0.15