Question

If a `2` `kg` object moving at `6` `ms^-1` slows down to a halt after moving `10` `m`, what is the friction coefficient of the surface that the object was moving over?

Answer 1

The object's initial kinetic energy will be used as work overcoming the frictional force. The calculation below shows that the coefficient of friction is approximately 5.4.

Explanation:

The object's initial kinetic energy is given by:

`E_k=1/2mv^2=1/2xx2xx6^2=36` `J`

This energy will be dissipated doing work against the frictional force acting on the object.

`W=Fd`

Rearranging:

`F=W/d=(36J)/(10m)=3.6` `N`

This is the magnitude of the frictional force, which is given by:

`F_"frict"=muF_N`, where `F_N` is the normal force, which in turn is the weight force of the object:

`F_N=mg=2*9.8=19.6` `N`

To find the frictional force, then:

`mu=(F_N)/(F_"frict")=19.6/3.6~~5.4`

(coefficients of friction do not have units)

(this is an improbable value for a coefficient of friction - they are typically less than 1, but occasionally as high as 2. Teachers and textbooks do sometimes write questions with these improbable values, though)