A 300-kg crate is placed on an adjustable inclined plane.As one end of the incline is raised,the crate begins to move downward.If the crate slides down the plane with an acceleration of 0.70 m/s² when the incline angle is 25 $\degree$,what is the coefficient of kinetic friction between ramp and crate? (g = 9.8 m/s²) A)0.47 B)0.42 C)0.39 D)0.12

Question

Quizplus · Accepted Answer

The net force down the incline is m*g*sin(θ) - friction, where friction = μ_k*m*g*cos(θ). The net force also equals m*a. Solving for μ_k gives μ_k = (g*sin(θ) - a) / (g*cos(θ)) = (9.8*sin(25°) - 0.70) / (9.8*cos(25°)) ≈ 0.39.

A 300-Kg Crate Is Placed on an Adjustable Inclined Plane °\degree°

A 300-Kg Crate Is Placed on an Adjustable Inclined Plane $\degree$