Block on Plane with Friction in strange coordinate system

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The discussion focuses on calculating the time T it takes for a block to come to rest while sliding down an inclined plane with friction. The user derives equations for the block's motion using a coordinate system where y is vertical and x is horizontal. They express the forces acting on the block, including gravitational and frictional forces, and attempt to find T by setting the velocity equations to zero. However, the user notes discrepancies between their derived answer and a previous solution obtained using a different coordinate system. They also mention issues with formatting their equations in LaTeX.
lordkelvin
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Homework Statement


A block of mass m slides down a plane inclined at an angle theta with initial velocity v down the slope and with friction coefficient mu. Find T, the time in which the block comes to a rest due to friction. Use coordinates with y vertical and x horizontal.

Homework Equations


F=ma
f=(mu)n

The Attempt at a Solution



\vec{v} = v \cos\theta \hat{x} - v \sin\theta \hat{y}

\\vec{\\F}_{w} = - m g \hat{y} \\

\vec{f} = - \mu mg \cos^{2}\theta \hat{x} + \mu m g \cos(\theta)\sin(\theta)\hat{y } \\

\ddot{x} = - \mu g \cos^{2}(\theta) \\

\dot{x} = - \mu g \cos^{2}(\theta) t + v\cos(\theta) \\

\ddot{y} = \mu g \cos(\theta)\sin(\theta) - g \\

\dot{y} = ( \mu g \cos(\theta)\sin(\theta) - g ) t - v \sin\theta \\

Now I should be able to set either y dot or x dot equal to zero and get T, right?
The answer I'm getting doesn't agree with the answer I got with the x-axis down the plane and the y-axis perpendicular to the plane. The answer I get in this way is

T = \frac {v_{0}}{g (\mu \cos \theta - \sin \theta)}

I hope this works; I've never used latex before.
 
Last edited:
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I can't get my latex fixed. Anyway, 'mu' should be a mu in the x dot equation, and the T equation shouldn't appear up top, instead it should say the force due to weight is minus mg in the y hat dir.

and v should end after the \\
 
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