# If a 4 kg object moving at 6 m/s slows to a halt after moving 35 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Feb 27, 2016

$\setminus {\mu}_{k} = 0.0525$

#### Explanation:

Since the object is moving on a plane surface, the kinetic friction experienced by the object will be $\setminus {\mu}_{k} m g$ implying that the de-acceleration due to friction will be $\setminus {\mu}_{k} g$

So, using the third kinematic equation
${v}^{2} = {v}_{o}^{2} + 2 a x$

The object is said to have come to a halt so that means $v = 0 m {s}^{-} 1$
So, ${v}_{o}^{2} = 2 a x$ (we ignored the negative sign since we know why that's there).

So, we know it travels a distance of $35 m$ before it stops so $x = 35 m$. We know it initially had a speed of $6 m {s}^{-} 1$ so ${v}_{o} = 6 m {s}^{-} 1$.

So, $a = {v}_{o}^{2} / \left(2 x\right) = {6}^{2} / \left(2 \cdot 35\right) = \frac{36}{70} = 0.51$
So, $a = 0.51$. But $a = \setminus {\mu}_{k} g$ So, $0.51 = \setminus {\mu}_{k} \cdot 9.8 \setminus \implies \setminus {\mu}_{k} = \frac{0.51}{9.8}$

Solving it yourself, you now understand that things look odd even if they are right.