Answer to Question #290616 in Mechanics | Relativity for mnzs

Question

A 2.00-kg block is given an initial speed of 2.50 m/s up a plane inclined 15.0° with the horizontal.

The coefficient of kinetic friction between block and plane is 0.250. What speed does it have just as

it returns to its initial position?

Answer 1

$\frac {mv^2}2=mgh+F_{fr}l,$

$\frac{v^2}2=gh+\frac{\mu gh}{\tan\alpha},$

$h=\frac{v^2\tan\alpha}{2g(\tan\alpha+\mu)};$

$mgh=\frac{mv_1^2}2+Fl,$

$gh=\frac{v_1^2}2+\frac{\mu gh}{\tan\alpha},$

$v_1=v\sqrt{\frac{\tan\alpha-\mu}{\tan\alpha+\mu}}=0.47~\frac ms.$