If a 2 kg object moving at 9 m/s slows down to a halt after moving 2 m, what is the friction coefficient of the surface that the object was moving over?

Aug 29, 2016

By conservation of energy the initial KE of the object must be same as the work done by it against the frictional force. So

$\frac{1}{2} m {v}^{2} = \mu m g \times x \ldots . \left(1\right)$

Where

$m \to \text{mass of the object}$

$\mu \to \text{coefficient of friction}$

$x \to \text{displacement}$

$g \to \text{Acceleration due to gravity}$

From (1)

$\mu = {v}^{2} / \left(2 \times g \times x\right)$

$\mu = {9}^{2} / \left(2 \cdot 9.8 .2\right) = \frac{81}{4 \cdot 9.8} = 2.06$

Sep 12, 2016

Alternate solution.

Explanation:

Using the kinematic equation
${v}^{2} - {u}^{2} = 2 a s$ where symbols have respective usual meanings, we get
${0}^{2} - {9}^{2} = 2 \times a \times 2$
$a = - \frac{81}{4} m {s}^{-} 2$
Decelerating force due to friction is $= m a$
$= 2 \times \left(- \frac{81}{4}\right) = - \frac{81}{2} N$

Equating it with Force due to friction $= - \mu m g$, we obtain
$\frac{81}{2} = \mu \times 2 \times 9.81$
Solving for $\mu$ we get
$\mu = \frac{\frac{81}{2}}{2 \times 9.81} = 2.06$, rounded to two decimal places.