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wcaprar

  • 4 years ago

An object of 180kg is sliding at a velocity of 3.78m/s. The object has a coefficient of kinetic friction of .70 as it slides. How far does the object slide?

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  1. fewscrewsmissing
    • 4 years ago
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    The kinetic force of friction is equal to the product of the normal force and the kinetic co-efficient of friction, or mg*μk. This allows you to calculate the acceleration (from F=ma) and the distance from basic kinematic equations: \[\Large \begin{array}{l} u = 3.78\,{\rm{m}}{{\rm{s}}^{ - 1}}\\ m = 180\,{\rm{kg}}\\ n = mg = 180 \times 9.81 = 1765.8\,{\rm{N}}\\ {\mu _k} = 0.70\\ {f_k} = {\mu _k}n = 0.70 \times 1765.8 = 1236.06\,{\rm{N}}\\ F = ma\\ a = \frac{F}{m} = \frac{{1236.06}}{{180}} = 6.867\,{\rm{m}}{{\rm{s}}^{ - 2}}\\ {v^2} = {u^2} + 2as\\ 0 = {3.78^2} + 2 \times 6.867 \times s\\ {3.78^2} = - 2 \times 6.867 \times s\\ s = \frac{{{{3.78}^2}}}{{2 \times 6.867}} = 1.04\,{\rm{m}} \end{array}\] It is very late and I am in the middle of exams so I may have misplaced a digit or two, but the concept's there =)

  2. wcaprar
    • 4 years ago
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    I am such an idiot! Lol! I have been staring at F=ma and thinking, I don't know the force and I need to find an acceleration... Where do I put force of friction?! Haha! Thanks, you really cleared it up

  3. fewscrewsmissing
    • 4 years ago
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    Haha it's alright, physics isn't one of those things that necessarily follows common sense or common thinking, the concepts have to be learnt. I'm sure I sat there looking at many a similar problem before I knew how to do it too :-P

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