A block of mass 0.244 kg is placed on top of a light, vertical spring of force constant 4 975 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise

As the spring returns to it's equilibrium position, it performs

1/2 (4975 N/m) (0.097 m)² ≈ 23 J

while the gravitational force (opposing the block's upward motion) performs

-(0.244 kg) g (0.097 m) ≈ -2.3 J

of work on the block. By the work energy theorem, the total work done on the block is equal to the change in its kinetic energy:

23 J - 2.3 J = 1/2 (0.244 kg) v² - 0

where v is the speed of the block at the moment it returns to the equilibrium position. Solve for v :

v² = (23 J - 2.3 J) / (1/2 (0.244 kg))

v = √((23 J - 2.3 J) / (1/2 (0.244 kg)))

v ≈ 44 m/s

After leaving the spring, block is in free fall, and at its maximum height h it has zero vertical velocity.

0² - (44 m/s)² = 2 (-g) h

Solve for h :

h = (44 m/s)² / (2g)

h ≈ 2.3 m

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.

i don't understand. explain it clearly