Question

If a `1 kg` object moving at `10 m/s` slows down to a halt after moving `50 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`mu=1/9.8`

Explanation:

First we must find the acceleration experienced by the object.
Here we will use the kinematic equation:
`(vf)^2=(vi)^2+2ad`
plug in our given data...
`0=100+100a`
`a=-1m/s^2`
We can now use this acceleration to calculate force using `F=ma`
`F=1kg*-1m/s^2`
`F=-1N` this must be the force of friction because no other resistive forces are given.

Kinetic friction is `fk=muN`
The normal force in this case is the weight of the object. `N=mg`
`N=1kg*9.8m/s^2`
`N=9.8`

Plugging in our values for `fk` and `N`....
`1=mu*9.8`
`mu=1/9.8`