Question

If an object is moving at `50 m/s` over a surface with a kinetic friction coefficient of `u_k=2 /g`, how far will the object continue to move?

Answer 1

`625m`

Explanation:

Because of kinetic frictional force,the object will undergo constant deceleration.

So,to find upto what length it will go,we can apply the relation between velocity (`v`),deceleration (`a`) and displacement (`s`)

i.e `v^2=u^2-2as`(where, `u` is the initial velocity)

Now, frictional force acting is `f=mumg=2/g×mg=2m` where, `m` is its mass.

So, deceleration i.e `a=f/m=2ms^-2`

Putting in the equation and also putting `v=0` as it will travel until its final velocity becomes zero.

So,we get, `0^2=50^2-2×2×s`

Or, `s=625m`

Answer 2

The distance is `=625m`

Explanation:

The coefficient of kinetic friction is

`mu_k=F_r/N`

The mass of the object is `=m`

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

The normal reaction is `N=mg`

Therefore,

`F_r=mu_kN=mu_kmg`

According to Newton's Second Law

`F=ma`

The acceleration is

`a=F/m=-F_r/m=-(mu_kmg)/m=-mu_kg`

The coefficient of kinetic friction is `mu_k=2/g`

`a=-2/g*g=-2ms^-2`

The initial velocity is `u=50ms^-1`

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

Apply the equation of motion

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

The distance is

`s=(v^2-u^2)/(2a)`

`=(0-50^2)/(2*-2)`

`=50^2/4`

`=625m`