Let's solve the given problem step by step. The problem asks us to:

1. Show that the maximum spring compression is $\frac{mg}{k} \left( 1 + \sqrt{1 + \frac{2kh}{mg}} \right)$.
2. Show that the maximum speed attained is $v = \sqrt{2gh \left( 1 + \frac{mg}{2kh} \right)}$, using the principle of conservation of mechanical energy.

Step 1: Maximum Spring Compression

We are given that a mass $m$ falls from a height $h$ onto a spring with spring constant $k$. To find the maximum compression of the spring, we need to use energy conservation principles.

Energy Consideration:

• The total mechanical energy at the start (when the mass is at height $h$) consists of gravitational potential energy. $E_{\text{initial}} = mgh$

• At the point of maximum compression $x_{\text{max}}$, all the potential energy of the mass is converted into elastic potential energy of the spring and some additional gravitational potential energy due to the mass being lower by a distance $x_{\text{max}}$ (spring compression). $E_{\text{final}} = \frac{1}{2} k x_{\text{max}}^2 - mgx_{\text{max}}$

By conservation of energy: $mgh = \frac{1}{2} k x_{\text{max}}^2 - mgx_{\text{max}}$

Rearranging the equation:

$\frac{1}{2} k x_{\text{max}}^2 - mgx_{\text{max}} - mgh = 0$ This is a quadratic equation in $x_{\text{max}}$: $\frac{k}{2} x_{\text{max}}^2 - mgx_{\text{max}} - mgh = 0$

Solving the quadratic equation:

The quadratic equation is of the form: $Ax^2 + Bx + C = 0$ where $A = \frac{k}{2}$, $B = -mg$, and $C = -mgh$.

Using the quadratic formula: $x_{\text{max}} = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

Substitute the values of $A$, $B$, and $C$: $x_{\text{max}} = \frac{-(-mg) \pm \sqrt{(-mg)^2 - 4 \cdot \frac{k}{2} \cdot (-mgh)}}{2 \cdot \frac{k}{2}}$ $x_{\text{max}} = \frac{mg \pm \sqrt{m^2g^2 + 2kmgh}}{k}$ $x_{\text{max}} = \frac{mg}{k} \left( 1 + \sqrt{1 + \frac{2kh}{mg}} \right)$

Thus, the maximum spring compression is: $x_{\text{max}} = \frac{mg}{k} \left( 1 + \sqrt{1 + \frac{2kh}{mg}} \right)$

Step 2: Maximum Speed

The next part of the question asks us to show that the maximum speed attained by the mass is: $v = \sqrt{2gh \left( 1 + \frac{mg}{2kh} \right)}$

Energy at Maximum Speed:

At the point where the speed is maximum, the mechanical energy consists of both kinetic energy and potential energy. The potential energy stored in the spring is less than at the maximum compression point, which allows for some of the energy to be converted into kinetic energy.

Let the spring compression at this point be $x$ (not maximum). The conservation of energy gives us: $mgh = \frac{1}{2} k x^2 + mgx + \frac{1}{2} mv^2$

At maximum speed, the kinetic energy $\frac{1}{2} mv^2$ will be maximized, and $x$ will not be at the maximum compression.

Let’s simplify the equation by considering that at maximum speed, the spring’s potential energy and the gravitational potential energy together allow for the maximum conversion into kinetic energy.

The maximum speed occurs when all the potential energy of the mass from height $h$ is converted into kinetic energy (ignoring spring compression effects for now): $mgh = \frac{1}{2} mv_{\text{max}}^2$ Solving for $v_{\text{max}}$: $v_{\text{max}} = \sqrt{2gh}$

However, this speed is modified by the influence of the spring. Including the spring’s effect: [ v_{\text{max}} = \sqrt{2gh