Question

A block is placed at the bottom of the incline and given an initial velocity Vo directed along the incline. The coefficient of the kinetic friction between the block and the surface is 'u'. Find what should be the angle of the incline 'A' so the block goes the minimal distance along the incline. Find that distance.

Answer 1

I know this was already a question here, but the methods for solution were poorly described.

Answer 2

u is constant as it is property of the surface( don't think of it as tan theta).
To get the lowest distance you need the largest force decreasing its velocity, so you need to get the first derivative and make it equal zero.
\[derivative (Mg \sin \theta + u Mg \cos \theta ) = 0\]
you have M, g and u constants
\[Mg \cos \theta - u Mg \sin \theta = 0\]
Mg cancelled
\[\cos \theta = u \sin \theta\]
\[\tan \theta = u\] \[\theta = \tan^{-1} \frac{ 1 }{ u }\]
And use this equation \[V _{t}^{2}- V _{0}^{2} = 2 a x\]
we said that\[F down ward = M a = Mg \sin \theta + u Mg \cos \theta \]
and get a isolated :D
\[a = g (\sin \theta + u \cos \theta)\]
finally get the answer\[X = \frac{ V _{0}^{2} }{ 2 g ( sintheta+ u \cos \theta)}\]

Answer 3

I forgot to mention http://openstudy.com/study#/updates/50162b5ee4b04dfc808a8424