A steel safe with mass M falls onto a concrete floor. Just before hitting the floor, its speed is vi . It hits the floor without rebounding, and ends up being d shorter than before. Find the magnitude of the (average) force exerted by the floor on the safe while it is being deformed. You can assume this force is a constant during the short time it takes for the safe to deform. Ignore the effects of gravity.

Explanation:

initial velocity, u = vi

final velocity, v = 0

distance = d

mass of safe = M

Use third equation of motion

v² = u² + 2as

0 = vi² - 2 ad

a = vi²/2d

Force, F = mass x acceleration

F = M x vi²/2d

F = Mvi²/2d

the second one and the forth one are in the purple because they are opinions and the rest are in the yellow

explanation:

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.