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wcaprar
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?
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 =)
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
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