If the coefficient of friction is 0.090. what is the ski’s speed at the base of the incline?

A ski starts from rest and slides down a 28 degree


Given that

The initial speed of the skier (u) =0m/s

The ski makes an anlge with the incline is (theta) =28degrees

The length if the incline is (s) =85m

The coefficient of friction is (uk) =0.090

The acceleration due to gravity (g) =9.81m/s2

If we consider the free-body diagram

The vertical force is given by

SigmaFy =N-mgcos28 =0

N =mgcos28

Now the height of the incline is given by

H =85sin28 =39.905m=40m

Now from the conservation of energy equation the work done is given by

mgh -85ukmgcos(28) =(1/2)mv2

gh -85ukgcos(28)=0.5v2

9.81*40-85*0.090*9.81*cos(28) =0.5v2



v =25.539m/s


The distance travelled is s =v2/2ukg =(25.539)2/2*0.090*9.81 =369.373m

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