Question

If a `2 kg` object moving at `9 m/s` slows down to a halt after moving `2 m`, what is the friction coefficient of the surface that the object was moving over?

Answer 1

By conservation of energy the initial KE of the object must be same as the work done by it against the frictional force. So

`1/2mv^2=mumgxxx....(1)`

Where

`m -> "mass of the object"`

`mu->"coefficient of friction"`

`x->"displacement"`

`g->"Acceleration due to gravity"`

From (1)

`mu=v^2/(2xxgxxx)`

`mu=9^2/(2*9.8.2)=81/(4*9.8)=2.06`

Answer 2

Alternate solution.

Explanation:

Using the kinematic equation
`v^2-u^2=2as` where symbols have respective usual meanings, we get
`0^2-9^2=2xxaxx2`
`a=-81/4ms^-2`
Decelerating force due to friction is `=ma`
`=2xx(-81/4)=-81/2N`

Equating it with Force due to friction `=-mumg`, we obtain
`81/2=muxx2xx9.81`
Solving for `mu` we get
`mu=(81/2)/(2xx9.81)=2.06`, rounded to two decimal places.