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.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#2.3">$\leftarrow$Back</a></h3>

    <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
$$F_y:~N=k\cdot \Delta x$$
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
$$\boxed{F=mg\left(1+\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)}$$

    <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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