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+ <meta name="description" content="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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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <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.
+</p>
+<center>
+ <figure>
+ <img src="2.3.45.png"
+ loading="lazy" width="170" />
+ <figcaption>
+ For problem $2.3.45$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="2.3.45_1.png"
+ loading="lazy" width="320" />
+ <figcaption>
+ Compression of the spring under the force of the weight's gravity
+ </figcaption>
+ </figure>
+</center>
+<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
+$$\boxed{F=mg\left(1+\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)}$$
+ </p>
+
+ <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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