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

Jan 28, 2016

By using kinematics, we can determine the acceleration, and use that to find the coefficient of friction.

#### Explanation:

By Newton's ${2}^{n d}$ law, we know that acceleration is the result of a force, in this case the force of friction.

${F}_{\text{net}} = m a$

To find acceleration, we have three variables:
${v}_{i} = 8 \frac{m}{s}$
${v}_{f} = 0 \frac{m}{s}$
$\Delta d = 4 m$

We can find acceleration using:

${v}_{f}^{2} = {v}_{i}^{2} + 2 a \Delta d$

${0}^{2} = {8}^{2} + 2 a \cdot 4$
$a = - {8}^{2} / \left(2 \cdot 4\right) = - 8 \frac{m}{s} ^ 2$

Now, with the acceleration, we could sub it in to the second law formula above, find the force of friction, and use that to find the coefficient, but let me show you a little magical shortcut that not only makes it faster, but also eliminates the need to know the mass of the object!

${F}_{\text{net}} = m a$

In this case, the net force is provided by friction, and friction is described by it's coefficient as follows:

${F}_{\text{friction" = mu_k F_"Normal}}$

and in this case, ${F}_{\text{Normal}} = m g$

so ${F}_{\text{net}} = {\mu}_{k} m g$

${\mu}_{k} m g = m a$ and the masses cancel out!

${\mu}_{k} \cancel{m} g = \cancel{m} a$ and

${\mu}_{k} = \frac{a}{g}$

If we use 10 for g (you may be using 9.8, or 9.81 in your class):

${\mu}_{k} = \frac{8}{10} = 0.8$