A ski starts from rest and slides down a 28 degree incline 85 m long. If the coefficient of friction is 0.090, what is the ski's speed at the base of the incline? 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? Use energy methods.

## Answer

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/s^{2}

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)=0.5v^{2}

9.81*40-85*0.090*9.81*cos(28) =0.5v^{2}

392.4-66.262=0.5v^{2}

326.138=0.5v^{2}

v =25.539m/s

b)

The distance travelled is s =v^{2}/2ukg
=(25.539)^{2}/2*0.090*9.81 =369.373m

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