A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s2. At the end of 2.0 s its angular momentum about the origin is:

hello your question is incomplete below is the complete question

A 2.0 kg block starts from rest on a positive x axis 3.0m from the origin and thereafter has an acceleration given by a= 4.0i - 3.0jin m/s2. At the end of 2.0 s, its angular momentum about the origin is .

answer : L = ( 2 kg ) [(-18m^2/s)k]

Explanation:

velocity vector(v)  = (4.0i - 3.0j )(2.0s)

position vector(r) = ( 3.0m)i

determine the angular momentum about the origin

L = m ( r * v )

attached below is the detailed solution

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.

it has to be b, about 97 percent.