Kinetics question 1

Question

A snowboarder takes speed downhill in order to get up a hill without pushing. If he arrives at the bottom of the hill with a velocity of 60 km/h, will he be able to get up a hill 75 m long with an 8% slope? (Kinetic friction coefficient: snowboard-snow = 0.1 and air resistance is neglected). (Hint: a 1% slope rises 1 m vertically for every 100 m of lateral travel.)

Answer 1

Let

v = 60 km/h = 50/3 m/s be his speed at the bottom,

s = 75 m be the length of the hill,

α be the angle of the hill,

tan(α) = 0.08,

f = 0.1 be the coefficient of friction,

m be his mass (including equipment),

g = 9.81 m/s² be gravitational acceleration,

h = sin(α)s be the height of the hill,

P = mgh be his potential energy at the top,

K = mv²/2 be his kinetic energy at the bottom,

F = mgcos(α) be the component of his weight normal to the slope,

W = Ffs be the work performed by friction on his way to the top.

To get up the hill, his kinetic energy at the bottom must be enough

for his potential energy at the top plus the work of friction.

K ≥ P + W

mv²/2 ≥ mgsin(α)s + mgcos(α)fs

v²/2 ≥ gsin(α)s + gcos(α)fs

v²/2 ≥ gs(tan(α) + f)cos(α)

v²/2gs(tan(α) + f) ≥ cos(α)

Substituting actual numbers

(50/3)²/(2×9.81×75×(0.08+0.1)) ≥ cos(α)

1.04 ≥ cos(α)

That inequality is always satisfied without having to calculate cos(α).

Therefore he will make it up the hill.