A body of mass m falls from a height h onto a spring of stiffness k and length l standing vertically on the floor. Determine the maximum pressure on the floor. Explain why this force increases as the spring stiffness increases.

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Solutions of Savchenko Problems in Physics
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    <h3> Statement </h3>
    <p>
        $2.3.45.$ A body of mass $m$ falls from a height $h$ onto a spring of stiffness $k$ and length $l$ standing vertically on the floor. Determine the maximum pressure on the floor. Explain why this force increases as the spring stiffness increases.

    <h3>Solution</h3>
    <p>

Conservation of kinetic and potential energies
$$mgh = \frac{mv^2}{2} + mgH + \frac{k(l-H)^2}{2}$$
Since at a finite instant of time, the body stops $(v=0)$
$$mgh=mg(l-\Delta x)+\frac{k\Delta x^2}{2}\quad (1)$$
From the drawing
$$H=l-\Delta x$$
Let's substitute into $(1)$
$$\frac{k\Delta x^2}{2}-mg\Delta x-mg(h-l)=0$$
Solve as a quadratic equation
$$\Delta x = \frac{mg}{k}\left(1\pm\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)$$
As $\Delta x$ must be positive, we keep only the plus solution
$$\Delta x = \frac{mg}{k}\left(1+\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)$$
Since at the bottom point, there are no forces other than the support reaction force $N$ and spring elasticity force $k\Delta x$, and at the bottom position the acceleration is zero ($a=0$) and the forces are compensated, the condition of equilibrium on the vertical axis has the form
$$F_y=N$$
Whence we obtain the maximum force of pressure on the floor, which is equal to the force with which the support acts on the load, according to Newton's third law
$$F_y=k\cdot \Delta x\Rightarrow \boxed{N=mg\left(1+\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)}$$
Analysing the obtained expression, we obtain that $F\sim \sqrt{k}\to$ Therefore, the greater the spring stiffness $k$, the greater the support reaction force $N$

    <h4>Answer</h4>
    <p>
        $$F=mg\left(1+\sqrt{1+2k(h-l)/(mg)}\right)$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
      Almaskhan Arsen<br>
    </p>


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