Answer
The proof is below.
Work Step by Step
We know that the force of friction is given by: $\mu_k F_n=mu_k mgcos\theta$
We use conservation of energy to find:
$mgh=\frac{1}{2}mv^2 - F_n \mu \Delta x$
$mgh=\frac{1}{2}mv^2 - mg \mu h\frac{1}{tan\theta}$
We know that at the bottom of the ramp:
$\frac{1}{2}m(\frac{1}{2}v)^2=mgh - mg \mu h\frac{1}{tan\theta}$
Using substitution, it follows:
$\frac{1}{2}m(\frac{1}{2}v)^2=\frac{1}{2}mv^2 - mg\mu h\frac{1}{tan\theta} - mg\mu h\frac{1}{tan\theta}$
$-\frac{3}{8}v^2 =- 2g \mu h\frac{1}{tan\theta} $
We need to find a value for h. Thus, we obatin:
$mgh =\frac{1}{2}mv^2 - mg \mu h\frac{1}{tan\theta} $
$h =\frac{1}{2g}v^2 - \mu h\frac{1}{tan\theta}$
$h + \mu h\frac{1}{tan\theta}=\frac{1}{2g}v^2 $
$h (1+ \mu \frac{1}{tan\theta})=\frac{v^2}{2g} $
$h = \frac{v^2}{2g(1+ \mu \frac{1}{tan\theta})}$
Using this, we find:
$-\frac{3}{8}v^2 =- 2g \mu (\frac{v^2}{2g(1+ \mu \frac{1}{tan\theta})})\frac{1}{tan\theta} $
$-\frac{3}{8}=\frac{-\mu \frac{1}{tan\theta}}{1+ \mu \frac{1}{tan\theta}}$
$(1+ \mu \frac{1}{tan\theta})(-\frac{3}{8})=-\mu \frac{1}{tan\theta}$
$\frac{-3}{8}=-\frac{5}{8} \mu \frac{1}{tan\theta}$
$\mu=\frac{3}{5}tan\theta$
Hence, the proof is completed.
