$2.3.45.$ 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.
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 $$\boxed{F=mg\left(1+\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)}$$
$$F=mg\left(1+\sqrt{1+2k(h-l)/(mg)}\right)$$
Almaskhan Arsen