Question

If you push an object along a surface with friction show that the distance it will travel depends on the angle of inclination of the surface θ as
S=v0^2/2g(sinθ+cosθµk)


where v0 is the initial speed and the µk is the coefficient of kinetic friction.

Answer 1

Well, here are the diagrams:We will use this kinematic equation:\[S=v _{o}t+\frac{ 1 }{ 2 } at ^{2}\]First we need to figure out the acceleration of the system "a." For that we need to find the net force. In the "y" direction, there is no acceleration, but in the "x" direction there is. So for the "x" direction:\[F _{x}=-Fr-mg _{x}=-Nu _{k}-mg \sin \theta =ma \rightarrow a=\frac{ -Nu _{k}-mg \sin \theta }{ m } \]But this is incomplete since we don't know what the normal force "N" is. To find it we need to go back to the "y" direction:\[N-mg _{y}=ma _{y}=0=N-mg \cos \theta \rightarrow N=mg \cos \theta \]Now we can finish our second equation and we get:\[a=\frac{ -mg \cos \theta u _{k}-mg \sin \theta }{ m }=-g(\cos \theta u _{k}+\sin \theta)\]Second, we need to find the time that the block took to travel "s". To do this we will solve for time from the definition of acceleration:\[a=\frac{ v _{f}-v_{o} }{ t } \rightarrow t=\frac{ v _{f}-v_{o} }{ a }\]And since the final velocity is zero then:\[t=-\frac{ v_{o} }{ a }\] Finally we just need to plug in what we got for "a" and "t":\[S=v _{o}\frac{ v _{o} }{ g(\cos \theta u _{k}+\sin \theta) }-\frac{ 1 }{ 2 }g(\cos \theta u _{k}+\sin \theta)(\frac{ v _{o} }{ g(\cos \theta u _{k}+\sin \theta) })^{2}\]Which if you simplify is:\[S=\frac{ v _{o}^{2} }{ 2g(\cos \theta u _{k}+\sin \theta) }\]

Answer 2

hmm..now i got it...thank you so much@ivancsc1996