Block on Plane with Friction in strange coordinate system

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SUMMARY

The discussion focuses on calculating the time T for a block of mass m to come to rest while sliding down an inclined plane at an angle theta, with an initial velocity v and a friction coefficient mu. The equations of motion are derived using vector notation, leading to expressions for acceleration in both the x and y directions. The final formula for T is established as T = v0 / (g (mu cos(theta) - sin(theta))), which is confirmed to be consistent when using the correct coordinate system. The user expresses confusion regarding the application of LaTeX and the proper setup of the equations.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Familiarity with friction forces (f=(mu)n)
  • Knowledge of vector decomposition in physics
  • Basic proficiency in LaTeX for mathematical notation
NEXT STEPS
  • Study the derivation of motion equations in inclined planes
  • Learn about the effects of friction on motion in different coordinate systems
  • Explore advanced applications of LaTeX for physics equations
  • Investigate the implications of varying the angle theta and friction coefficient mu on the time T
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined planes, as well as educators looking for examples of motion with friction in non-standard coordinate systems.

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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