Posted by **emmie** on Tuesday, May 8, 2012 at 1:20pm.

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 .

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 .

- physics -
**Elena**, Tuesday, May 8, 2012 at 2:05pm
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.

- physics -
**Anonymous**, Tuesday, October 21, 2014 at 10:46am
a=3.05

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