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A body is slid upwards through an inclined plane of angle [tex]\theta=15\degree[/tex]. Determine the friction coefficient of the body with the surface of the plane if the time it takes to go upwards [tex]t_a[/tex] is double the time it takes to go downwards [tex]t_b[/tex] ([tex]t_a=2t_b[/tex]).

Mi procedure is as follows:

The normal force: [tex]N=mg\cos(\theta)[/tex].

1. Upwards:

During the ascent the x componen of the weight and the friction force act in the same direction, therefore: [tex]a_a=-g\sin(\theta)-\mu g\cos(\theta)=-g(\sin(\theta)+\mu\cos(\theta))[/tex].

The ecuacion of space as a function of time is: [tex]x=2v_0 t_b -2g(\sin(\theta)+\mu\cos(\theta))t_b^2[/tex] (1).

The ecuation of speed as a function of time is: [tex]0=v_0 -2g(\sin(\theta)+\mu\cos(\theta))t_b[/tex] (2).

2. Downwards

During the descent the x component of the weight and the friction force act in opposite directions, therefore: [tex]a_a=-g\sin(\theta)+\mu g\cos(\theta)=-g(\sin(\theta)-\mu\cos(\theta))[/tex].

The ecuation of space as a function of time is: [tex]0=x-\frac{1}{2}g(\sin(\theta)-\mu\cos(\theta))t_b^2[/tex] (3).

Combining (1), (2) and (3) we obtain:

[tex]\frac{1}{2}g(\sin(\theta)-\mu\cos(\theta))t_b^2=4g(\sin(\theta)+\mu\cos(\theta))t_b^2-2g(\sin(\theta)+\mu\cos(\theta))t_b^2[/tex]

[tex]\sin(\theta)-\mu\cos(\theta)=4\sin(\theta)+4\mu\cos(\theta)[/tex]

[tex]\mu=-\frac{3}{5}\tan(\theta)[/tex]

From here I deduce that

**a value for [tex]\mu[/tex], for with the conditions are met, doesn't exist.**Is it correct?

Thanks! :)