FOR A liquid drop of radius r falling in air with terminal velocity, power delivered by gravitational force is propotional to

P = m g v

Explanation:

Power is defined as the relationship between work and time

        P = W / t

The work is defined by the scalar product of the force and the displacement, in the case of a drop falling the vertical force and the displacement has the same direction, therefore the angle between them is zero and the cosines of zero is one (cos 0 = 1)

       W = F y

the gravitational force is proportional to the weight of the drop  (F = mg)

        W = mg y

we substitute

       P = m g y / t

the terminal velocity of the drop is constant

       P = m g v

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

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.