Calculating Distance Traveled by a Skier on an Inclined Slope

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To calculate the distance a skier travels down a slope inclined at 4.7 degrees before coming to rest, one must consider the initial speed of 2.7 m/s and the coefficient of kinetic friction of 0.11. The net force acting on the skier includes gravitational forces and friction, which can be expressed through equations involving mass and acceleration. The skier will decelerate due to friction, and the equation Vf^2 - Vi^2 = 2a * distance can be applied to find the stopping distance. It is crucial to determine whether the skier's applied force affects the stopping distance, as the frictional force may be sufficient to slow the skier without additional effort. The calculations should focus on these forces to accurately determine the distance traveled before coming to a stop.
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The Question
A skier on a slope inclined at 4.7 degrees to the horizontal pushes on ski poles and starts down the slope. The initial speed is 2.7m/s. The coefficient of kinetic friction between skies and snow is 0.11. Determine how far the skier will slide before coming to rest.

So far i had done the following, but i am unsure if this is the right way to go about this and also what to do next to find the answer.

in the perpendicular
0 = mgcos4.7 [D] + Fn
Fn = mgcos4.7

Ff = 0.11(mgcos4.7)


parallel
Fnet = Ff + Fgperp + fapp
0 = mgsin4.7 - 0.11(mgcos4.7) - Fapp
Fapp = 0.27(m)
 
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Little bit confused, where does it say a force is applied by the skiier to help stop.

if no effort, along incline ma=mg(sin 4.7)-mg(cos(4.7))*.11

If cos(4.7)*0.11 is greater than sin(4.7), skiier will slow without any effort.

One could use that number a, to compute a and relation

Vf^2-Vi^2=2a*distance
 
Of course - I had forgotton that equation - thank you
 
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