A 75kg snowboarder has an initial velocity of 5.0 m/s at the top of a 28 degree incline. After sliding down the 110m long incline (on which the coefficient of kinetic friction is = Bk0.18), the snowboarder has attained a velocity v . The snowboarder then slides along a flat surface (on which Bk= 0.15) and comes to rest after a distance x .

Question

Part a) Use Newton's second law to find the snowboarder's acceleration while on the incline and while on the flat surface.

part b)Then use these accelerations to determine x .

Answer 1

m =75 kg, vₒ=5 m/s, s = 110 m,
k1 = 0.18, k2 = 0.15.
a1=? a2=?, x =?

The equations of the motion along the incline are
x: m•a1 = m•g•sinα – F1(fr),
y: 0 = - m•g•cosα + N.
F1(fr) = k1•N=k1•m•g•cos α,
m•a1 = m•g•sinα – k1•m•g•cos α,
a1 = g• (sinα - k1•cos α) = 9.8•(0.469 – 0.18•0.883) = 3.04 m/s².

v=vₒ +a1•t,
s =vₒ•t + a1•t²/2 .
v = sqrt(2•a1•s +vₒ²) = sqrt(2•3.04•110 + 25) = 26.33 m/s.

The motion along a flat surface:
m•a2 = F2(fr) = k2•m•g,
a2 = k2•g = 0.15•9.8 = 1.47 m/s².
x = v²/2•a2 = (26.33)²/2•1.47 = 235 m.

Answer 2

a=3.05