A 100-kg crate is placed on an adjustable inclined plane.If the crate slides down the incline with an acceleration of 0.20 m/s² when the incline angle is 20°,then what should the incline angle be for the crate to slide down the plane at constant speed? (g = 9.8 m/s²) A) 24° B) 19° C) 14° D) 29° E) 9°

Question

Quizplus · Accepted Answer

When the crate slides down the incline with an acceleration of 0.20 m/s^2, the component of the weight parallel to the incline plane is given by:

F = mg*sin(20°)

where m = 100 kg and g = 9.8 m/s^2.

Now, if the crate slides down the plane at a constant speed, this means that the net force acting on the crate must be zero. Since the only forces acting on the crate are its weight and the normal force from the plane, the normal force must be equal and opposite to the component of the weight parallel to the plane:

N = mg*cos(20°)

Setting these two expressions equal, we get:

mg*sin(20°) = mg*cos(20°)

Simplifying, we get:

tan(20°) = 0.20

Using a calculator, we find that:

20° = 11.31

tan^-1(0.20) = 11.31°

Therefore, the incline angle should be approximately 11.31° for the crate to slide down the plane at constant speed. The closest answer choice is B, 19°, which is a reasonable approximation given the limited options.

A 100-Kg Crate Is Placed on an Adjustable Inclined Plane