# If a 10kg object moving at 12 ms^-1 slows down to a halt after moving 7 m, what is the friction coefficient of the surface that the object was moving over?

Jan 25, 2016

First step is to find the acceleration, then the frictional force, then compare it to the normal force to find the friction coefficient.

#### Explanation:

First we find the acceleration. The initial velocity, $u$, is $12$ $m {s}^{-} 1$, the final velocity, $v$ is $0$ $m {s}^{-} 1$, and the distance is $7$ $m$.

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

Rearranging:

$a = \frac{{v}^{2} - {u}^{2}}{2 d} = \frac{{0}^{2} - {12}^{2}}{2 \cdot 7} = - 10.3$ $m {s}^{-} 2$

The negative sign just shows that this is a deceleration (acceleration in the opposite direction to the velocity), so we can ignore it from now on.

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

$F = m a = 10 \cdot 10.3 = 103$ $N$

This is the frictional force, and the frictional coefficient compares this with the 'normal force': in this case the weight force of the object:

$\mu = {F}_{f} / {F}_{N} = \frac{103}{10 \cdot 9.8} = 1.05$

Two things here:

1. $\mu$, the friction coefficient, is a dimensionless number - it has no units.

2. This answer is impossible. The friction coefficient is always between 0 and 1. A friction coefficient greater than 1 cannot exist, since it means the frictional force is greater than the normal force.

I think I've done all the calculations correctly, so it looks to me as though there is an error in the question, and the numbers given do not lead to a physically possible answer.

It's very important not to just do the calculations and give back the number the equations spit out, but to think about the physics of the answer, and what it means .