If a $9$ $kg$ object moving at $9$ $ms^-1$ slows down to a halt after moving $81$ $m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2017-12-04T07:43:59

The frictional coefficient $mu=0.102$

First step is to find the acceleration:

$v^2=u^2+2as$

$a=(v^2-u^2)/(2s)=(0^2-9^2)/81=-1$ $ms^-2$

(the negative sign just indicates that the acceleration is in the opposite direction to the initial velocity)

Second step is to find the force causing the acceleration:

$F=ma=9xx-1=-9$ $N$

This is the magnitude of the frictional force. The third step is to find the normal force, which is just the weight force on the object:

$F=mg=9xx9.8=88.2$ $N$

The final step is to find the friction coefficient:

$mu=F_"fric"/F_"norm"=9/88.2=0.102$

Frictional coefficients do not have units.

If a 99 kgkg object moving at 99 ms^-1ms^-1 slows down to a halt after moving 8181 mm, what is the coefficient of kinetic friction of the surface that the object was moving over?