A 150-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 0.40 m/s² when the incline angle is 25°,what is the coefficient of kinetic friction between ramp and crate? (g = 9.8 m/s²) A) 0.52 B) 0.62 C) 0.42 D) 0.32 E) 3.74

Question

Quizplus · Accepted Answer

We can use the following formula for the acceleration of the crate on the inclined plane:

a = g(sinθ - μcosθ)

where a is the acceleration of the crate, g is the acceleration due to gravity, θ is the angle of the incline, and μ is the coefficient of kinetic friction.

Substituting the given values:

0.40 m/s2 = 9.8 m/s2 (sin25° - μcos25°)

Solving for μ:

μ = (9.8 m/s2 sin25° - 0.40 m/s2)/ (9.8 m/s2 cos25°)

μ ≈ 0.42

Therefore, the coefficient of kinetic friction between the ramp and the crate is approximately 0.42. The correct answer is C.

A 150-Kg Crate Is Placed on an Adjustable Inclined Plane