Question

You've gone off on a ski trip to Lake Tahoe. You’re standing on the top of a hill, and then you push off and slide down a slope inclined at 20◦ from the horizontal ground. You ski 100 meters down the hill, until the ground eventually becomes level, and then you continue skiing another 248 meters along the flat ground before the friction eventually stops you.

(a) If the friction is always the same during your entire ski run, what is the coefficient of kinetic friction, µk, between your skis and the snow?

(b) What is your velocity at the bottom of the hill, before you slide horizontally?

Answer 1

a.)

By applying energy conservtion between top of hill and final point,

KEi + PEi + Wfr = KEf + PEf

here, KEi = initial kinetic energy = 0

PEi = initial potential energy = m*g*h

Wfr = work done by friction = fri*di*cosA + frh*dh*cosA

KEf = final kinetic energy = 0

PEf = final potential energy = 0

here, d = total displacement = di + dh = 100 + 248 = 348 m

A = angle between friction and displacement = 180 deg

h = height of hill = 100*sin

= 20.0 deg

fri = friction force acting on inclined plane = *(m*g*cos)

frh = friction force on horizontal plane =  *m*g

So, 0 + m*g*(100*sin) + *(m*g*cos)*di*cosA + *m*g*dh*cosA = 0 + 0

Using known values:

100*sin(20.0 deg) = *(100*cos(20.0 deg) + 248)

= (100*sin(20.0 deg))/[*(100*cos(20.0 deg) + 248)]

= 0.10 = Coefficient of kinetic friction

b.)

By applying energy conservtion between top of hill and bottom of hill

KEi + PEi + Wfr = KEf + PEf

here, KEi = initial kinetic energy = 0

PEi = initial potential energy = m*g*h

Wfr = work done by friction = fri*di*cosA

KEf = final kinetic energy = 0.5*m*v^2

PEf = final potential energy = 0

here, d = total displacement = di = 100 m

A = angle between friction and displacement = 180 deg

h = height of hill = 100*sin

= 20.0 deg

fri = friction force acting on inclined plane = *(m*g*cos)

So, 0 + m*g*100*sin + *(m*g*cos)*di*cosA = 0.5*m*v^2 + 0

using known values:

v^2 = 2*[(9.81*100*sin(20.0 deg)) + 0.10*9.81*cos(20.0 deg)*100*cos(180 deg)]

v = sqrt[2*(9.81*100*sin(20.0 deg)) + 2*0.10*9.81*cos(20.0 deg)*100*cos(180 deg)] = 22.06

v = 22 m/s = velocity at the bottom