Forces - time to stop

  • Thread starter dvyu
  • Start date
  • #1
34
0
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)
 

Answers and Replies

  • #2
961
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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
 
  • #3
34
0
Of course - I had forgotton that equation - thank you
 

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