If a 3 kg object moving at 30 ms^-1 slows to a halt after moving 9 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Feb 28, 2016

There are a couple of possible approaches to beginning this question: I have briefly outlined both below. The later stages of the solution are the same. The frictional coefficient calculated by either method is $5.1$.

Explanation:

Method 1: starting by finding the force using the acceleration

$u = 30$ $m {s}^{-} 1$
$v = 0$ $m {s}^{-} 1$
$s = 9$ $m$ (some people use $d$ for the distance)
a = ? $m {s}^{-} 2$

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

Rearranging:

$a = \frac{{v}^{2} - {u}^{2}}{2 s} = \frac{0 - {30}^{2}}{2 \times 9} = - \frac{900}{18} \approx - 50$ $m {s}^{-} 2$

The minus sign just means the acceleration is in the opposite direction to the initial velocity, which makes sense because the object slows down. Understanding the direction, we don't need to use the minus sign in the rest of our calculations.

Knowing the mass, we can use Newton's Second Law to find the force acting, which is the frictional force:

$F = m a = 3 \cdot 50 = 150$ $N$

From this point onward the approach is the same in both methods:

The frictional force is related to the normal force by the expression:

${F}_{\text{fric"=muF_"norm}}$

The normal force is just the weight force of the object, $m g$, so, rearranging:

$\mu = {F}_{\text{fric"/F_"norm"=F_"fric}} / \left(m g\right) = \frac{150}{3 \cdot 9.8} = 5.1$

Frictional coefficients relate forces in $N$, and have no units themselves.

Method 2: starting by finding the force from the work done

The work done to stop a moving object is equal to the change in kinetic energy of the object. This one has $0$ $J$ at the end and

${E}_{k} = \frac{1}{2} m {v}^{2} = \frac{1}{2} \cdot 3 \cdot {30}^{2} = 1350$ $J$ at the start.

The work done is given by $W = F d$ and we can rearrange this to find the frictional force:

$F = \frac{W}{d} = \frac{1350}{9} = 150$ $N$

Now we know the frictional force - and we're very happy that it's the same as the force we calculated using a different method! We can go through the rest of the steps in the same way and calculate the coefficient of friction.

(Side note: $5.1$ is not a very realistic value for a coefficient of friction: these are typically between $0$ and $1$, sometimes as high as $2$ but rarely much higher than that. That's just an issue with how the question was set.)

Feb 28, 2016

Here initial kinetic energy of the body = work done by the body against the frictional force
$\frac{1}{2} m {v}^{2} = \mu m g \cdot d$
$\implies {v}^{2} = 2 \cdot \mu g \cdot d$
$\implies \mu = {v}^{2} / \left(2 g d\right) = {30}^{2} / \left(2 \cdot 10 \cdot 9\right) = 5$

the answer is impossible one since $\mu$cannot be greater than 1