If a 3/2 kg object moving at 3/4 ms^-1 slows to a halt after moving 5/8 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Aug 1, 2016

First find the acceleration, then use it to find the magnitude of the frictional force, and use that to find the coefficient of friction.

Explanation:

First find the acceleration:

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

Rearranging to make $a$ the subject:

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

$a = \frac{{0}^{2} - {\left(\frac{3}{4}\right)}^{2}}{2 \times \left(\frac{5}{8}\right)} = - 0.45$ $m {s}^{-} 2$

The negative sign just shows that the object is decelerating - its acceleration is in the opposite direction to the initial velocity.

Now to find the frictional force:

$F = m a = \left(\frac{3}{2}\right) \times \left(0.45\right) = 0.675$ $N$

The frictional force is given by ${F}_{\text{frict" = muF_"norm}}$, where the normal force is ${F}_{\text{norm}} = m g$.

Rearranging to make the frictional coefficient the subject:

$\mu = \left({F}_{\text{frict")/(F_"norm}}\right) = \frac{0.675}{\frac{3}{2} \times 9.8} = 0.022$

(frictional coefficients have no units, since they are ratios of forces)

(note: technically, a coefficient of friction is between two bodies, not a property of the surface alone)