A 75-kg snowboarder has an initial velocity of 5.0 m/s at the top of a 28 ∘ incline. after sliding down the 110-m long incline (on which the coefficient of kinetic friction is μk = 0.18), the snowboarder has attained a velocity v. the snowboarder then slides along a flat surface (on which μk = 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.

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1 November, 22:13

A 75-kg snowboarder has an initial velocity of 5.0 m/s at the top of a 28 ∘ incline. after sliding down the 110-m long incline (on which the coefficient of kinetic friction is μk = 0.18), the snowboarder has attained a velocity v. the snowboarder then slides along a flat surface (on which μk = 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.

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Janiyah Weaver · Answer 1

Let the following denote:m - mass of the snowboarder, g - acceleration due to gravity, F - friction force, a - acceleration down the slope, R - normal reaction of the slope, u - coefficient of friction on the slope, v0 - snowboarder's initial speed, v - snowboarder's speed at the base of the slope, F1 - friction force on the level, a1 - acceleration on the level,

Resolving parallel and perpendicular to the slope: mg sin (a) - F = ma would be equation 1 mg cos (a) = R would be equation 2F = uR would be equation 3

From (2) and (3) : F = umg cos (a).

Substituting this in (1) : mg[ sin (a) - u cos (a) ] = ma a = g[ sin (a) - u cos (a) ] = 9.81[ sin (28) - 0.18 cos (28) ] = 3.05 m/s^2.

v^2 = v0^2 + 2as v = sqrt (5.0^2 + 2 * 3.05 * 110) = 26.4 m/s.

F1 = - u1 mg - u1 mg = m a1 a1 = - u1 g = - 0.15 * 9.81 = - 1.47 m/s^2