A skier slides horizontally along the snow for a distance of 11.9 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.0458. Initially, how fast was the skier going?

Question

A skier slides horizontally along the snow for a distance of 11.9 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.0458. Initially, how fast was the skier going?

I do not know how to start this. Thank you.

Physics(Please help) - Elena, Saturday, June 2, 2012 at 8:59am
ΔKE =KE2 –KE1 = 0 -m•v²/2.
W(fr) =μ•m•g•s•cosα,
where α is the angle between the friction force and displacement.
α =180º, cos α = -1.
-m•v²/2 = - μ•m•g•s,
v = sqrt(2•μ• g•s).

So for v would I do sqrt(2*11.9*9.8*0.0458) ?

Answer 1

I got 10.68 as my answer but my homework said this was incorrect.Why?

Answer 2

v = sqrt(2•μ• g•s)= sqrt(2•0.0458•9.8•11.9) = 3.27 m/s

Answer 3

Yes, you're on the right track! To find the initial velocity (v), you can use the equation v = sqrt(2 * μ * g * s), where μ is the coefficient of kinetic friction, g is the acceleration due to gravity (approximately 9.8 m/s^2), and s is the distance traveled before coming to rest (11.9 m in this case).

So, plugging in the given values into the equation, you would have v = sqrt(2 * 0.0458 * 9.8 * 11.9). Evaluating this expression, you would find the initial velocity of the skier.