If a 2 kg object moving at 10 ms^-1 slows to a halt after moving 25 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Jun 17, 2017

$\setminus \mu = {F}_{\text{frict}} / \left(m g\right) = \frac{4}{2 \times 9.8} = 0.20$

Explanation:

The object's final velocity will be $v = 0$ $m {s}^{-} 1$, the initial velocity $u = 10$ $m {s}^{-} 1$ and the distance $25$ $m$. We can use this to find the object's acceleration (deceleration):

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

$a = \frac{{v}^{2} - {u}^{2}}{2 s} = \frac{{0}^{1} - {10}^{2}}{2 \times 25} = \frac{- 100}{50} = - 2$ $m {s}^{-} 2$

The frictional force will be given by ${F}_{\text{frict}} = m a = 2 \times - 2 = - 4$ $N$

That is, the force is $4$ $N$ in the direction opposite to the initial motion.

The relationship between the frictional force and the object's weight force is ${F}_{\text{frict}} = \setminus \mu {F}_{w} = \setminus \mu m g$.

Rearranging:

$\setminus \mu = {F}_{\text{frict}} / \left(m g\right) = \frac{4}{2 \times 9.8} = 0.20$

(frictional coefficients have no units)