A 2.0-kg block slides on a frictionless 25$\degree$ inclined plane.A force of 4.6 N acting parallel to the incline and up the incline is applied to the block.What is the acceleration of the block? A)1.8 m/s² up the incline B)2.3 m/s² up the incline C)6.6 m/s² down the incline D)1.8 m/s² down the incline E)2.3 m/s² down the incline

Question

Quizplus · Accepted Answer

The force parallel to the incline is 4.6 N. We need to resolve this force into its components, one parallel to the incline and one perpendicular to the incline. The force perpendicular to the incline is canceled out by the normal force provided by the inclined plane because there is no friction between the block and the inclined plane. The force parallel to the incline produces an acceleration in the direction of the force.

We can use the formula F = ma, where F = 4.6 N and m = 2.0 kg. The acceleration a is in the direction of the force parallel to the incline, which is up the incline.

The component of the gravitational force parallel to the incline is mg sin(25F1F1F1S1F1F1F10), where g is the acceleration due to gravity (9.8 m/s^2).

We can use Newton's second law in the direction parallel to the incline:

F_parallel - mg sin(25F1F1F1S1F1F1F10) = ma

Substituting the values, we get

4.6 N - 2.0 kg x 9.8 m/s^2 x sin(25F1F1F1S1F1F1F10) = 2.0 kg x a

a = 1.8 m/s^2 up the incline

Therefore, the correct answer is D, 1.8 m/s^2 down the incline.

A 20-Kg Block Slides on a Frictionless 25 °\degree° Inclined Plane

A 20-Kg Block Slides on a Frictionless 25 $\degree$ Inclined Plane