If an object is moving at $12 m/s$ over a surface with a kinetic friction coefficient of $u_k=5 /g$ , how far will the object continue to move?

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Answer 1 · 2017-04-02T06:40:50

The distance is $=14.4m$

Let the mass of the object be $=mkg$

$v=12ms^-1$

The coefficient of kinetic friction is

$mu_k=F_r/N$

$N=mg$

So,

$F_r=mu_k*N$

$=5/g*mg=5m N$

The work done by $F_r$ is equal to the kinetic energy of the object.

$F_r*d=1/2mv^2$

$d=(1/2mv^2)/(5m)=v^2/10=12^2/10=144/10=14.4m$

If an object is moving at 12 m/s12 m/s over a surface with a kinetic friction coefficient of u_k=5 /gu_k=5 /g, how far will the object continue to move?