A block is given an initial velocity of 3.00 m/s up a frictionless incline of angle 20 degrees. How far up the incline does the block slide

The distance moved by the block is 1.34 m

Explanation:

Given;

initial velocity of the block, u = 3 m/s

angle of inclination, θ = 20⁰

The net horizontal force acting on the block;

∑Fx = - mgsinθ

Apply Newton's second law of motion, to determine the constant acceleration of the block.

∑Fx = ma

- mgsinθ = ma

-gsinθ = a

Apply the following Kinematic equation, to determine how far up, the block moved before coming to rest.

Therefore, the distance moved by the block is 1.34 m

positive reinforcement? ?

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.