# If a 4 kg object moving at 5/4 m/s slows to a halt after moving 8 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Jun 26, 2016

$\approx 0.01$

#### Explanation:

Given
$m \to \text{Mass of the object} = 4 k g$

$m \to \text{Velocity of the object} = \frac{5}{4} \frac{m}{s}$
$g \to \text{Acceleration due to gravity} = 9.8 \frac{m}{s} ^ 2$

$d \to \text{Distance travelled before halt} = 8 m$

mu_k->"Coefficient of kinetic friction"=?

Considering law of conservation of energy we can say,
$\text{work done against friction =KE of the object}$

$\implies {\mu}_{k} m g \times d = \frac{1}{2} \cdot m {v}^{2}$

=>mu_k=v^2/(2*g*d)=(5/4)^2/(2*9 .8*8)~~0.01