A block of mass m is dropped from a height h onto a spring with spring constant K. What is the maximum distance the spring will be compressed?

Let $m$ be the mass of the block, $h$ be the initial height above the spring, $K$ be the spring constant, $g$ be the acceleration due to gravity ($9.8 \text{ m/s}^2$), and $x$ be the maximum compression of the spring.

Initial state (block at height $h$ above the uncompressed spring):
The block has gravitational potential energy. We set the reference point for potential energy at the maximum compression of the spring. So, the total initial height of the block relative to this lowest point is $h+x$.
Initial potential energy ($PE_{initial}$) = $mg(h+x)$
Initial kinetic energy ($KE_{initial}$) = 0 (since it's dropped)
Total initial energy ($E_{initial}$) = $mg(h+x)$

Final state (spring maximally compressed by $x$):
The block is momentarily at rest, so $KE_{final} = 0$.
The spring is compressed by $x$, so it has elastic potential energy ($PE_{elastic}$) = $\frac{1}{2}Kx^2$.
At this point, the block is at the reference height for gravitational potential energy, so $PE_{gravitational, final} = 0$.
Total final energy ($E_{final}$) = $\frac{1}{2}Kx^2$

$$E_{initial} = E_{final}$$ $$mg(h+x) = \frac{1}{2}Kx^2$$

Given:
$m = 4 \text{ kg}$
$h = 0.80 \text{ m}$
$K = 1916 \text{ N/m}$
$g = 9.8 \text{ m/s}^2$

$$4 \text{ kg} \times 9.8 \text{ m/s}^2 \times (0.80 \text{ m} + x) = \frac{1}{2} \times 1916 \text{ N/m} \times x^2$$ $$39.2 (0.80 + x) = 958 x^2$$ $$31.36 + 39.2x = 958x^2$$

Here, $a = 958$, $b = -39.2$, and $c = -31.36$.

$$x = \frac{-(-39.2) \pm \sqrt{(-39.2)^2 - 4(958)(-31.36)}}{2(958)}$$ $$x = \frac{39.2 \pm \sqrt{1536.64 + 120000.64}}{1916}$$ $$x = \frac{39.2 \pm \sqrt{121537.28}}{1916}$$ $$x = \frac{39.2 \pm 348.622}{1916}$$

We get two possible values for $x$:

$$x_1 = \frac{39.2 + 348.622}{1916} = \frac{387.822}{1916} \approx 0.2024 \text{ m}$$ $$x_2 = \frac{39.2 - 348.622}{1916} = \frac{-309.422}{1916} \approx -0.1615 \text{ m}$$

Since $x$ represents a distance of compression, it must be a positive value.
Therefore, $x \approx 0.2024 \text{ m}$.

Rounding to three significant figures:

$$x \approx 0.202 \text{ m}$$

The maximum distance through which the spring will be compressed is:

$$0.202 \text{ m}$$