If a 11 kg object moving at 15 ms^-1 slows down to a halt after moving 225 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Jan 19, 2016

The coefficient of friction is related to the force acting on the object to accelerate (decelerate) it and the normal force: $\mu = \frac{F}{F} _ N = \frac{5.5}{107.8} = 0.05$.

Explanation:

First we need to calculate the acceleration (deceleration) of the object:

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

${0}^{2} = {15}^{2} - 2 \cdot a \cdot 225$

$a = 0.5 m {s}^{-} 2$

Now we have an acceleration and a mass and want to find a force... hmm, that sounds a lot like Newton's Second Law to me!

$F = m a = 11 \cdot 0.5 = 5.5 N$

This is the actual magnitude of the frictional force. The coefficient of friction is defined as:

$\mu = \frac{F}{F} _ N$ where $\mu$ is the coefficient and ${F}_{N}$ is the normal force acting. In this case that is the weight force of the object, given by ${F}_{N} = m g = 11 \cdot 9.8 = 107.8 N$

Then $\mu = \frac{F}{F} _ N = \frac{5.5}{107.8} = 0.05$

Note that a frictional coefficient is a 'dimensionless' number which does not have units (because it is calculated as a force in $N$ divided by another force in $N$, so the units cancel).