A wedge of mass [tex]m = 37.1 \, \text{kg}[/tex] is located on a plane that is inclined by an angle [tex]\theta = 20.5^\circ[/tex] with respect to the horizontal. A force [tex]F = 311.3 \, \text{N}[/tex] in the horizontal direction pushes on the wedge, as shown. The coefficient of friction between the wedge and the plane is [tex]0.153[/tex].

What is the acceleration of the wedge along the plane? (Use negative numbers for motion to the left and positive numbers for motion to the right.)

The acceleration of the wedge along the inclined plane, considering gravitational, normal, frictional, and applied forces, is approximately 4.76 m/s^2.

To determine the acceleration of the wedge along the inclined plane, Newton's second law is applied, stating that the net force on an object is the product of its mass and acceleration. Considering the forces acting on the wedge:

Gravitational force pulling the wedge downward along the inclined plane.

Normal force perpendicular to the plane.

Frictional force opposing motion along the plane.

Horizontal force applied externally.

First, resolve forces perpendicular and parallel to the inclined plane:

Perpendicular to the plane: N = mg * cos(theta), where N is the normal force, m is the mass of the wedge, g is the acceleration due to gravity, and theta is the angle of inclination.

Parallel to the plane: F_parallel = mg * sin(theta), force parallel to the plane.

Calculate frictional force using the coefficient of friction (mu):

Frictional force: f_friction = mu * N

Now, determine the net force along the plane:

Net force: F_net = F - f_friction = ma

Substitute expressions for F_parallel and f_friction into the net force equation:

a = (F - mu * mg * cos(theta)) / m

Now, plug in the given values:

a = (311.3 - 0.153 * 37.1 * 9.81 * cos(20.5)) / 37.1

a ≈ (311.3 - 0.153 * 37.1 * 9.81 * 0.9397) / 37.1

a ≈ (311.3 - 134.49) / 37.1

a ≈ 176.81 / 37.1

a ≈ 4.76 m/s^2

Therefore, the acceleration of the wedge along the plane is approximately 4.76 m/s^2.