# If an object with a mass of 10 kg  is moving on a surface at 15 ms^-1 and slows to a halt after  4 s, what is the friction coefficient of the surface?

Jan 20, 2016

The acceleration of the object is $3.75 m {s}^{-} 2$, which means the frictional force acting is $37.5 N$ and the frictional coefficient is $\mu = 0.38$.

#### Explanation:

First we calculate the acceleration of the object:

$v = u + a t$

where $v$ is final velocity $\left(m {s}^{-} 1\right)$, $u$ is initial velocity 9ms^-1), $a$ is the acceleration $\left(m {s}^{-} 2\right)$ and $t$ is the time $\left(s\right)$.

Rearranging:

$a = \frac{v - u}{t} = \frac{0 - 15}{4} = - 3.75 m {s}^{-} 2$

The negative sign just shows that it is a deceleration, and we can ignore it for the rest of this calculation.

Now we find the force acting to slow the object, using Newton's Second Law :

$F = m a = 10 \cdot 3.75 = 37.5 N$

This is simply the frictional force acting. Frictional force is related to the normal force acting on the object, which in this case is the weight of the object, ${F}_{N} = m g = 10 \cdot 9.8 = 98 N$:

${F}_{f} = \mu {F}_{N}$

Rearranging and substituting in the force:

$\mu = {F}_{f} / {F}_{N} = \frac{37.5}{98} = 0.38$

Note that $\mu$ is a dimensionless number, that is, it doesn't have any units.