Calculating Distance Traveled by a Skier on an Inclined Slope

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SUMMARY

The discussion focuses on calculating the distance a skier travels down a slope inclined at 4.7 degrees with an initial speed of 2.7 m/s and a coefficient of kinetic friction of 0.11. The skier's motion is analyzed using forces acting parallel and perpendicular to the slope, leading to the conclusion that if the frictional force exceeds the gravitational component down the slope, the skier will decelerate without additional effort. The relevant equations include the net force equation and the kinematic equation Vf² - Vi² = 2a * distance, which are essential for determining the stopping distance.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with kinematic equations
  • Knowledge of forces acting on inclined planes
  • Basic grasp of friction and its coefficients
NEXT STEPS
  • Study the derivation of forces on inclined planes in physics
  • Learn about kinematic equations in detail, particularly Vf² - Vi² = 2a * distance
  • Explore the effects of varying coefficients of friction on motion
  • Investigate real-world applications of these principles in sports physics
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Physics students, educators, and anyone interested in understanding the dynamics of motion on inclined surfaces, particularly in sports contexts like skiing.

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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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