Question

A skier traveling 12.0 m/s reaches the foot of a steady upward 18.0º incline and glides 12.2 m up along this slope before coming to rest. What was the average coefficient of friction?

Answer 1

Answer:

$\mu = 0.31$

Explanation:

As we know that skier start from the foot of the inclined plane with speed

v = 12 m/s

the angle of the inclined plane is

$\theta = 18 degree$

length of the inclined plane = 12.2 m

final speed of the skier = 0

so here we can use energy conservation

Work done against gravity + work done against friction = change in kinetic energy

so we will have

$-mgLsin\theta - \mu mg Lcos\theta = 0 - \frac{1}{2}mv^2$

$(9.81)(12.2)sin18 + (\mu)(9.81)(12.2)cos(18) = \frac{1}{2}(12^2)$

$36.98 + \mu(113.8) = 72$

$\mu = 0.31$