A 75g person on skis is going down a hill sloped at an angle 30.the coefficient of kinetic friction between the skis and the snow is 0.15.how fast is the skier going 10s after starting from rest?

Answer 1

First determine the acceleration, a.
The net force down the slope equals M*a. The net force is the gravity component down the slope minus the friction force.
M*g*sin30 - M*g*cos30*0.15 = M*a

M cancels out, and so
a = g*(sinA -0.15 cosA) = 0.37 g
= 3.6 m/s^2

Then use V = a*t for the velocity at time t.