A block of mass m has a spring –spring constant k and length L –connected, vertically, to the bottom.

The block is dropped from height h, towards the floor. Assuming the spring is stiff enough that it does not
compress fully at the floor, what is the closest distance, z, the block gets to the floor? (Note that the quadratic
equation might seem ugly but things like k, L, h, m, g are all given so there’s really only one unknown in
there.)

mgh=kx²/2

x=sqrt{2mgh/k}

Z=L- sqrt{2mgh/k}

To find the closest distance, z, the block gets to the floor, we can use the principle of conservation of mechanical energy.

At the initial height h, the block has potential energy which is equal to mgh (mass * gravity * height).

At the closest distance z, the block will have transformed all its potential energy into elastic potential energy in the compressed spring and gravitational potential energy at height z.

The potential energy stored in the compressed spring is given by (1/2)kz^2 (spring constant * compression distance squared).

Therefore, the potential energy at distance z is (1/2)kz^2 + mgh.

Since energy is conserved, the initial potential energy at height h is equal to the potential energy at distance z:

mgh = (1/2)kz^2 + mgh.

Rearranging the equation, we get:

(1/2)kz^2 = mgh - mgh
(1/2)kz^2 = 0
z^2 = 0
z = 0.

Therefore, the closest distance the block gets to the floor is 0.