Question

If a `8 kg` object moving at `16 m/s` slows to a halt after moving `80 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

0.16

Explanation:

Clearly,here the frictional force caused retardation of the moving object and brought it to rest.

So,using `v^2 = u^2 - 2as` (all symbols are bearing their conventional meaning)(given, `v=0` ,`u=16 m/s` and `s = 80`)

We get, `a = 1.6 m/s^2`

Now,if the coefficient of kinetic friction is `mu` , frictional force(`f`) acting on the body was `mu*N` or `mu*mg` i.e `mu*80`

So,we using `f = ma` (where, `m=8Kg` and `a = 1.6 m/s^2`)

We, get `mu*80 = 12.8`
Or, `mu = 0.16`

Answer 2

The coefficient of kinetic friction is `=0.16`

Explanation:

The mass of the object is `m=8kg`

The acceleration due to gravity is `g=9.8ms^-2`

The Normal reaction is

`N=mg=8xx9.8=78.4N`

The initial velocity is `u=16ms^-1`

The final velocity is `v=0ms^-1`

The distance is `s=80m`

Apply the equation of motion,

`v^2=u^2+2as`

The acceleration is

`a=(v^2-u^2)/(2s)=(0^2-16^2)/(2*80)=-1.6ms^-2`

According to Newton's Second Law

`F=ma`

The force of friction is

`F_r=8*1.6=12.8N`

The coefficient of kinetic friction is

`mu_k=F_r/N=12.8/78.4=0.16`