Question

A ski starts from rest and slides down a 20° incline 90 m long. Use energy methods.

(a) If the coefficient of friction is 0.09, what is the ski's speed at the base of the incline? m/s

(b) If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level? m

Answer 1

a) FN - mgcos28 = 0
=> FN = mgcos28

So, FR = u*FN = 0.09mgcos28

Initial Energy = KE0 + PE0
= 0 + mgh
= (mg)(90sin28)

FInal Energy = KEf + PEf
= 1/2*mv^2 + 0

Work done = FRdcos(theta) = 0.09mgcos28*90*cos180

So,
1/2mv^2 = (mg)(90sin28) - 0.09mgcos28*90*cos180
0.5*v^2 = (90sin28) - 0.09*9.8*cos28*90*cos180
=> v = 15 m/s

b) Work done by friction = (0.09mg)*dcos180 = -(0.09mg)*d

Initial Energy = 1/2mv^2 + 0 = 0.5*m*14.98^2

FInal Energy = 0

Final Energy = Initial Energy + Work done by friction

=> 0 = 0.5*m*14.98^2-(0.09mg)*d
=> d = 127.2 m