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
a) the time taken for the skier to come to a stop
b) the distance travelled down the hill
component of weight down hill = m g sin 5
friction force up hill = 0.2 m g cos 5
net force down hill = m g ( sin 5 - 0.2 cos 5) = m g(.0872-0.199)= -0.112 mg
so we stop
m a = -0.112m g
a = - .112*9.81 = - 1.1 m/s^2
v = Vi + a t = 3.5 - 1.1 t
so v = 0 when t = 3.5/1.1 = 3.18 seconds
d = Vi t - (1/2) a t^2
or just use average speed * t
= (3.5/2)*3.18 = 5.57 meterrs