Question

please show how to get the mew k ( or coefficient) on the lefthand side to solve

Answer 1

Answer:$\mu_k=\text{ }\frac{(0.5v^2-gh)}{gd\cos\theta}$

Explanations:

From the last line of the equation shown in the picture:

$-\mu_kmgd\cos \theta\text{ = }\frac{1}{2}mv^2-\text{mgh}$

This can be further simplified as:

$\begin{gathered} -\mu_kmgd\cos \theta\text{ =}0.5mv^2-\text{mgh} \\ -\mu_kmgd\cos \theta\text{ =m}(0.5v^2-gh) \\ \text{Divide both sides by -mgd}\cos \theta \\ \frac{-\mu_kmgd\cos\theta\text{ }}{-\text{mgd}\cos\theta}=\text{ }\frac{\text{m(}0.5v^2-gh)}{-\text{mgd}\cos\theta} \end{gathered}$

The expression representing μk (Coeffiecient of kinetic friction) is therefore written as:

$\begin{gathered} \mu_k=\text{ }\frac{\text{m(}0.5v^2-gh)}{-\text{mgd}\cos\theta} \\ \mu_k=\text{ }\frac{(0.5v^2-gh)}{gd\cos\theta} \end{gathered}$

There