A61 kg skier on level snow coasts 184 m to a stop from a speed of 12 m/s. (a) use the work-energy theorem to find the coefficient of kinetic friction between the skis and snow. (b) suppose a 75 kg skier with twice the starting speed coasted the same distance before stopping. find the coefficient of kinetic friction between that skier’s skis and the snow.

a) μ = 0.0398

b) μ = 0.1595

Explanation:

a) d = 184 m

Vi = 12 m/s

vf = 0 m/s

m = 61 Kg

μ = ?

We use the equation

W = Kf - Ki  (I)

then we have

W = F(friction)*d = - μ*N*d = - μ*m*g*d  (II)

Kf = m*(vf)²/2 = m*(0)²/2 = 0 J

If (I) = (II)

0 - m*(vi)²/2 = - μ*m*g*d    ⇒   μ = (vi)² / (2*g*d)

⇒   μ = (12 m/s)² / (2*9.81 m/s²*184 m) = 0.0398

b) d = 184 m

Vi = 24 m/s

vf = 0 m/s

m = 75 Kg

μ = ?

We use the equation

W = Kf - Ki  (I)

then we have

W = F(friction)*d = - μ*N*d = - μ*m*g*d  (II)

Kf = m*(vf)²/2 = m*(0)²/2 = 0 J

If (I) = (II)

0 - m*(vi)²/2 = - μ*m*g*d    ⇒   μ = (vi)² / (2*g*d)

⇒   μ = (24 m/s)² / (2*9.81 m/s²*184 m) = 0.1595

The mass of the skier is not important in the analysis.

unbalanced force

explanation: