Question

A snowboarder starts from rest on an inclined planar snow surface and gives herself a tiny push to start sliding down the hill. The coefficient of kinetic friction between her snowboard and the snow is 0.08. She notices that she travels 30.11 m horizontally and 37.18 m vertically (downward) before whizzing by a marker flag. How fast is she traveling as she passes the flag?

Answer 1

from the given data

angle of inclination, theta = tan^-1(y/x)

= tan^-1(37.18/30.11)

= 51.0 degrees

let m is the mass of snowboarder

net force acting on it along the surface, Fnet = m*g*sin(theta) - fk

m*a = m*g*sin(theta) - mue_k*N

m*a = m*g*sin(theta) - mue_k*m*g*cos(theta)

a = g*sin(theta) - mue_k*g*cos(theta)

= 9.8*sin(51) - 0.08*9.8*cos(51)

= 7.123 m/s^2

distance travelled along the surface, d = sqrt(x^2 + y^2)

= sqrt(30.11^2 + 37.18^2)

= 47.84 m

now use, vf^2 -vi^2 = 2*a*d

vf^2 - 0^2 = 2*a*d

vf = sqrt(2*a*d)

= sqrt(2*7.123*47.84)

= 26.11 m/s <<<<<<<<<<<<<--------------------Answer