Answer to Question #273946 in Mechanics | Relativity for Rosyid

Question

the 50-kg, uniform crate slides down the inclined surface with an initial speed 10 m/s. if the coefficient of kinetic friction between the crate and the inclined surface is 0.15, determine the distance s the crate has moved when it momentarily stops. (Assume tipping over does not happen)

Answer 1

$l=v^2/(2a)$

$a=g\sin30°-\mu g\cos30°$

We have

$l=\frac{v^2}{2\cdot(g\sin30°-\mu g\cos30°)}=\frac{10^2}{2\cdot(9.8\cdot \sin30°-0.15\cdot9.8\cdot\cos30°)}=13.8\ (m)$ . Answer