(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I am running to school at 6m/s when I hit a patch of icy sidealk. The coefficient of friction between the ice and my shoes is 0.16. I wish to turn my 55kg frame to the left. What is the smallest turn I can make?

v = 6 m/s

μ = 0.16

m = 55 kg

2. Relevant equations

Fnet = mv^2/r

Ff = μFn

Fnet = ma

3. The attempt at a solution

Assume that up and left are positive.

Fnet = ma

Fg + Fn = ma

-539 + Fn = 0

Fn = 539

Ff = μFn

Ff = 0.16 (539)

Ff = 86.24

Fnet = mv^2/r

Ff = mv^2/r

86.24 = 55(6)^2/r

86.24 = 1980/r

r = 23 m

I'm not sure if that's the answer or what to do next, or even if what I did is the right method. The question is kind of confusing me because I don't fully understand how the person is moving.

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# Centripetal Force/Changing direction on ice - Turn length

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