Question

If a `13 kg` object moving at `7 m/s` 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

`mu_k=0.25`

Explanation:

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

1. Method 1 - Using Newton's Laws and Equations of Motion :

The acceleration of the object is uniform and in 1 direction and can be found from a relevant equation of motion for constant acceleration in 1 dimension as follows :

`v^2=u^2+2ax`

`therefore a=(v^2-u^2)/(2x)=(0^2-7^2)/(2xx10)=-2.45m//s^2`.

This is acceleration is caused by the frictional force which is the resultant force acting on the object and so by Newton 2 and definition of frictional forces we get :

`sumvecF=mveca`

`therefore -f_k=ma`, where `f_k=mu_kN`.

But `N=mg =>mu_kmg=ma`.

`therefore mu_k=a/g=2.45/9.8=0.25`.

2. Method 2 - Using energy considerations :

The Work-Energy Theorem states that the work done by the resultant force equals the change in kinetic energy brought about.

`therefore W_(f_k)=DeltaE_k`

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

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

`therefore mu_k=(v^2-u^2)/(2gx)=(0^2-7^2)/(2xx9.8xx10)=0.25`.