A proton moving to the left at 10m⋅s−1 enters a magnetic field of strength 5.0×10−2T directed into the page. The proton is deflected into a circular path. During half a revolution, how much work is done on the proton by the magnetic force?

zero.

Explanation:

The work done by the magnetic force on the proton is zero. This is because direction of the magnetic force, which is towards the center of the circular path, is perpendicular to to the displacement of the proton, which is tangent to the path.

In other words,

,

and since the angle between the magnetic force and the displacement is 90°

.

This is a general result, true for all magnetic forces: The work done by a magnetic force is zero.

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.

resonant frequency