Question

A skier of mass 60 kg skies down a 35° slope; the coefficient of friction between her skis and the snow is 0.40

(a) Calculate the skier’s acceleration.
(b) If the skier started at 5 m/s, how fast is she going after travelling 25 m down the slope?

Answer 1

mu = 0.40, good grief wax those skis !!!!
normal force on snow = m g cos 35
so friction force up slope = mu m g cos 35=0.4*60*9.81*0.819 = 193 N
weight component down slope = m g sin 35 = 338 N
F = m a
338 - 193 = 60 a
a = 2.41 m/s^2
v = Vi + a t
v = 5 + 2.41 t
d = Vi t +(1/2) a t^2
d = 5 t + 1.2 t^2 = 25
1.2 t^2 + 5 t - 25 = 0
t =2.93 seconds
then v = 5 + 2.41 * 2.93 = 12.1 m/s

Answer 2

Bless you anonymous, thank you so much!