A sight seen on many bunny hills across Ontario is young skiers pushing on ski poles and gliding down a slope until they come to rest. Observing from a distance, you note a young person (approximately 25 kg) pushing off with the ski poles to give herself an initial velocity of 3.5 m/s. If the inclination of the hill is 5.0 degrees and the coefficient of kinetic friction for the skis on dry snow is 0.20, calculate

Question

a) the time taken for the skier to come to a stop
b) the distance travelled down the hill

Answer 1

component of weight down slope = m g sin 5
normal force = m g cos 5 so friction up slope = .2 m g cos 5
F = net force down slope = m g (sin 5 - .2 cos 5)
= m a
so a down slope = g(sin 5 - .2 cos 5) = g (-0.112)
if g = 9.81
then a = - 1.1 m/s^2
now the problem
v = Vi - 1.1 t = 3.5 -1.1 t
when v =0
t = 3.5/1.1 = 3.18 seconds
distance = average speed * t = (3.5/2) * 3.18 = 5.57 meters

Answer 2

thank you Damon