- #1
Lannie
- 11
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I'm trying to work out a problem involving the stopping distance of a skiier, and I thought it was a straightforward problem but I've gone wrong somewhere.
Determine the stopping distance for a skier moving down a slope with friction with an initial speed of 22.2 m/s.
Assume μk=0.173 and θ=4.9
To solve, I tried the formula:
Xinitial= Vsquared/2g(sinθ + μkcosθ)
so i used the data given and got
X=(22.2)squared/2(9.8)(sin4.9+0.173cos4.9)
= 97.5 m
but this did not work out.
Anyone have any thoughts?
Determine the stopping distance for a skier moving down a slope with friction with an initial speed of 22.2 m/s.
Assume μk=0.173 and θ=4.9
To solve, I tried the formula:
Xinitial= Vsquared/2g(sinθ + μkcosθ)
so i used the data given and got
X=(22.2)squared/2(9.8)(sin4.9+0.173cos4.9)
= 97.5 m
but this did not work out.
Anyone have any thoughts?