Question

What is the coefficient of kinetic friction if a skier accelerates down a 17 degree incline at a rate of 3.3 m/s^2? draw a diagram with all the forces.

Answer 1

Let's designate: $\mu$ - coefficient of kinetic friction,

$Ff - friction$ force,

$Fr$ - equally effective force,

$a = 3.3m / s^2$

$$\left\{ \begin{array}{c} F r = M a \\ F r = F t - F f \end{array} \right.$$ $$F n = F \cos \alpha$$ $$F t = F \sin \alpha$$ $$F f = \mu F n$$ $$M a = F (\sin \alpha - \mu \cos \alpha)$$ $$M a = M g (\sin \alpha - \mu \cos \alpha)$$ $$a = g \sin \alpha - \mu g \cos \alpha$$ $$\mu g \cos \alpha = a - g \sin \alpha$$ $$\mu = \frac {a - g \sin \alpha}{g \cos \alpha}$$ $$\mu = \frac {3 . 3 - 9 . 8 \times \sin (1 7 ^ {0})}{9 . 8 \times \cos (1 7 ^ {0})} = 0. 0 5$$

Answer: $\mu = 0.05$