# If a 13 kg object moving at 7 m/s slows down to a halt after moving 10 m, what is the friction coefficient of the surface that the object was moving over?

Jan 9, 2016

${\mu}_{k} = 0.25$

#### Explanation:

I will show you 2 different methods to do this question :

1. Method 1 - Using Newton's Laws and Equations of Motion :

The acceleration of the object is uniform and in 1 direction and can be found from a relevant equation of motion for constant acceleration in 1 dimension as follows :

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

$\therefore a = \frac{{v}^{2} - {u}^{2}}{2 x} = \frac{{0}^{2} - {7}^{2}}{2 \times 10} = - 2.45 m / {s}^{2}$.

This is acceleration is caused by the frictional force which is the resultant force acting on the object and so by Newton 2 and definition of frictional forces we get :

$\sum \vec{F} = m \vec{a}$

$\therefore - {f}_{k} = m a$, where ${f}_{k} = {\mu}_{k} N$.

But $N = m g \implies {\mu}_{k} m g = m a$.

$\therefore {\mu}_{k} = \frac{a}{g} = \frac{2.45}{9.8} = 0.25$.

2. Method 2 - Using energy considerations :

The Work-Energy Theorem states that the work done by the resultant force equals the change in kinetic energy brought about.

$\therefore {W}_{{f}_{k}} = \Delta {E}_{k}$

$\therefore {f}_{k} \cdot x = \frac{1}{2} m \left({v}^{2} - {u}^{2}\right)$

$\therefore {\mu}_{k} m g \cdot x = \frac{1}{2} m \left({v}^{2} - {u}^{2}\right)$

$\therefore {\mu}_{k} = \frac{{v}^{2} - {u}^{2}}{2 g x} = \frac{{0}^{2} - {7}^{2}}{2 \times 9.8 \times 10} = 0.25$.