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 0.20 m/s² when the incline angle is 18°,what is the coefficient of kinetic friction between ramp and crate? (g = 9.8 m/s²) A)0.40 B)0.50 C)0.30 D)0.20 E)2.8

Question

Quizplus · Accepted Answer

First, we need to find the force acting on the crate down the incline:

Fnet = ma
Fnet = 150 kg * 0.20 m/s^2
Fnet = 30 N

Next, we can find the component of the force of gravity acting down the incline:

Fg_parallel = mg*sin(18°)
Fg_parallel = 150 kg * 9.8 m/s^2 * sin(18°)
Fg_parallel = 488.3 N

Finally, we can find the coefficient of kinetic friction using the equation:

Ffriction = friction_coefficient * Fg_parallel

Solving for the friction coefficient:

friction_coefficient = Ffriction / Fg_parallel
friction_coefficient = Fnet / Fg_parallel
friction_coefficient = 30 N / 488.3 N
friction_coefficient = 0.0615

Therefore, the coefficient of kinetic friction between the ramp and crate is approximately 0.062 or 0.30, which matches choice C.

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