Question

If a `3 kg` object moving at `5 m/s` slows to a halt after moving `10 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

`mu_k=0,12755`

Explanation:

I will show you 2 different methods to do this question :

Method 1 - Using Newton's Laws and equations of motion

The acceleration of the object may be found from the equations of motion for constant linear acceleration as follows :
`v^2=u^2+2ax`
`thereforea=(v^2-u^2)/(2x)=(0^2-5^2)/(2xx10)=-1,25m//s^2`

This acceleration is caused by the resultant force of friction acting on the object and by Newton 2 given as :
`sumF=ma`
`therefore -f_k=ma`
`therefore -mu_kN=ma`
`therefore -mu_kmg=ma`

`therefore mu_k=(ma)/(-mg)=(-1,25)/(-9,8)=0,12755`.

Method 2 - Using energy considerations

From conservation of energy, work done by friction equals change in kinetic energy brought about.

`therefore W_ ( fk ) = Delta E_K`

`therefore f_k*x=1/2m(v^2-u^2)`

`therefore mu_kmg*x=1/2mv^2`

`therefore mu_k=(1/2mv^2)/(mgx)=(1/2xx5^2)/(9,8xx10)=0,12755`.