Question

Determine the stopping distance for a skier moving down a slope with friction with an initial speed of 19.0 m/s . Assume that θ = 5.62 deg, g=9.81 m/s2, and that the coefficient of kinetic friction between the skier and the slope is 0.289. Suppose the temperature drops by 2 degrees C and the skier waxes her skis; the coefficient of kinetic friction drops to 0.173. What is now the stopping distance, for the same initial velocity?

Answer 1

here,

theta = 5.62 degree

the accelration , a = (g * sin(theta) - uk * g * cos(theta) )

a = ( 9.81 * sin(5.62) - 0.289 * 9.81 * cos(5.62)) = - 1.86 m/s^2

initial speed , u = 19 m/s

the stopping distance for the skier , s = u^2 /2a

s = 19^2 /( 2*1.86) = 97 m

when the temprature drops by 2 degree C

uk = 0.173

the accelration , a = (g * sin(theta) - uk * g * cos(theta) )

a = ( 9.81 * sin(5.62) - 0.173 * 9.81 * cos(5.62)) = - 0.73 m/s^2

initial speed , u = 19 m/s

the stopping distance for the skier , s = u^2 /2a

s = 19^2 /( 2*0.73) = 247.8 m