Answer to Question #273946 in Mechanics | Relativity for Rosyid

Question #273946

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)


1
Expert's answer
2021-12-02T10:08:51-0500

l=v2/(2a)l=v^2/(2a)


a=gsin30°μgcos30°a=g\sin30°-\mu g\cos30°


We have


l=v22(gsin30°μgcos30°)=1022(9.8sin30°0.159.8cos30°)=13.8 (m)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

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