If an object is moving at $2$ $ms^-1$ over a surface with a kinetic friction coefficient of $mu_k=8/g$ , how far will the object continue to move?

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Answer 1 · 2016-02-20T10:20:16

The object will move $1/4$ (or 0.25) $m$ before it stops.

The frictional force will be given by:

$F_"frict"=mu*F_N$ where $F_N$ is the normal force, which is the weight force of the object: $F_N=mg$ .

$F_"frict"=mu*F_N=8/g*mg=8m$

The acceleration of the object will be described by Newton's Second Law :

$a=F/m=(8m)/m=8$ $ms^-2$

From its initial velocity, $u=2.0$ $ms^-1$ , to $v=0$ $ms^-1$ , the distance traveled will be given by:

$v^2=u^2+2ad$

Rearranging:

$d=(v^2-u^2)/(2a)=(0-2^2)/(2*8)=4/16=1/4$ $m$ .

If an object is moving at 22 ms^-1ms^-1 over a surface with a kinetic friction coefficient of mu_k=8/gmu_k=8/g, how far will the object continue to move?