# 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?

Aug 11, 2018

$625 m$

#### 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 ${v}^{2} = {u}^{2} - 2 a s$(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 $a = \frac{f}{m} = 2 m {s}^{-} 2$

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, $s = 625 m$

Aug 11, 2018

The distance is $= 625 m$

#### 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 $g = 9.8 m {s}^{-} 2$

The normal reaction is $N = m g$

Therefore,

${F}_{r} = {\mu}_{k} N = {\mu}_{k} m g$

According to Newton's Second Law

$F = m a$

The acceleration is

$a = \frac{F}{m} = - {F}_{r} / m = - \frac{{\mu}_{k} m g}{m} = - {\mu}_{k} g$

The coefficient of kinetic friction is ${\mu}_{k} = \frac{2}{g}$

$a = - \frac{2}{g} \cdot g = - 2 m {s}^{-} 2$

The initial velocity is $u = 50 m {s}^{-} 1$

The final velocity is $v = 0 m {s}^{-} 1$

Apply the equation of motion

${v}^{2} = {u}^{2} + 2 a s$

The distance is

$s = \frac{{v}^{2} - {u}^{2}}{2 a}$

$= \frac{0 - {50}^{2}}{2 \cdot - 2}$

$= {50}^{2} / 4$

$= 625 m$