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a skater with an initial speed of 7.60m/s is gliding across the ice. air resistance is negligible. (a)the coefficient of kinetic friction between the ice and the skate blades is 0.100. find the deceleration caused by kinetic friction. (b)how far will the skater travel before coming to rest?

someone had showed me something like this:
If the kinetic friction coeffient is 0.1, the force that is opposig motion is (F = 0.1 * M g).

(a) The deceleration rate is therefore
a = F/M = 0.1 g = 0.98 m/s^2

(b) The time to come to rest, T, is given by
a T = 7.6 m/s.
Solve for T

but i don't really understand how did he get the part : a = F/M = 0.1 g = 0.98 m/s^2

can anyone please give me some hints to do it or explain it to me? THANKS A LOT!

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5 answers
  1. Since Newton's second law says F = Ma, than a = F/m.
    g is the acceleration of gravity, 9.8 m/s^2

    I hope that explains what I wrote earlier

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  2. so what do you mean by 0.1g?

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  3. 0.1 g is g/10
    I got it by using the formul a = F/M
    I had already shown you that F = 0.1 M g

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  4. for b) the question is how FAR will it travel not long... so the answer is 29.4m

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  5. F=ma mew(Fn) N=W
    a=f/m
    a=mew(Fn)/m mew=coefficient of friction
    a=mew(W)/m
    a=mew(mg)/m
    a= mew(g)
    a=(0.100)(9.8)
    a=0.980m/s^2

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