Translated 2.1.34-2.1.47
en/2.1.36.md
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| + | <meta name="author" content="Aliaksandr Melnichenka"> | ||
| + | <meta name="date" content="2023-10" scheme="YYYY-MM"> | ||
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| + | <title>The velocity of a body of mass m in a viscous liquid decreases with the distance l traveled according to the law v = v_0 - \beta l, where v_0 is the initial velocity, and \beta is a constant coefficient. How does the viscous friction force acting on a body from the fluid side depend on the velocity of the body?</title> | ||
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| + | <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> | ||
| + | </div> | ||
| + | <p class="author"> | ||
| + | Solutions of Savchenko Problems in Physics <br> | ||
| + | <i><b>knowledge must be free</b></i> | ||
| + | </p> | ||
| + | </header> | ||
| + | |||
| + | <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $2.1.36.$ The velocity of a body of mass $m$ in a viscous liquid decreases with the distance $l$ traveled according to the law $v = v_0 - \beta l$, where $v_0$ is the initial velocity, and $\beta$ is a constant coefficient. How does the viscous friction force acting on a body from the fluid side depend on the velocity of the body? | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | The equation of Newton's second law for the direction of motion: | ||
| + | $$ ma = F_с $$ | ||
| + | |||
| + | $$ \frac{F_с}{m}=\frac{dv_x}{dt} $$ | ||
| + | Derivative of velocity with respect to time | ||
| + | $$ \frac{dv_x}{dt}=\frac{d}{dt}(v_0-βx) $$ | ||
| + | |||
| + | $$ \frac{dv_x}{dt}=\frac{dv_0}{dt}-β\frac{dx}{dt}=-βt $$ | ||
| + | the minus sign shows that the acceleration vector is directed in the direction opposite to the velocity vector. | ||
| + | Combining the equations, we obtain the value of the resistance force as a function of velocity | ||
| + | $$ \boxed{F=βmv} $$ | ||
| + | |||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$F = βmv$$ | ||
| + | </p> | ||
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| <meta charset="utf-8"> | |||
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| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
| <meta property="og:title" content="The velocity of a body of mass m in a viscous liquid decreases with the distance l traveled according to the law v = v_0 - \beta l, where v_0 is the initial velocity, and \beta is a constant coefficient. How does the viscous friction force acting on a body from the fluid side depend on the velocity of the body?"> | |||
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| <meta property="og:description" content="The velocity of a body of mass m in a viscous liquid decreases with the distance l traveled according to the law v = v_0 - \beta l, where v_0 is the initial velocity, and \beta is a constant coefficient. How does the viscous friction force acting on a body from the fluid side depend on the velocity of the body?"> | |||
| <meta name="yandex-verification" content="6cfda41f74038368"> | |||
| <title>The velocity of a body of mass m in a viscous liquid decreases with the distance l traveled according to the law v = v_0 - \beta l, where v_0 is the initial velocity, and \beta is a constant coefficient. How does the viscous friction force acting on a body from the fluid side depend on the velocity of the body?</title> | |||
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| </head> | |||
| <body style=""> | |||
| <header style="text-align:center;"> | |||
| <div id = "logo"> | |||
| <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> | |||
| </div> | |||
| <p class="author"> | |||
| Solutions of Savchenko Problems in Physics <br> | |||
| <i><b>knowledge must be free</b></i> | |||
| </p> | |||
| </header> | |||
| <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> | |||
| <h3> Statement </h3> | |||
| <p> | |||
| $2.1.36.$ The velocity of a body of mass $m$ in a viscous liquid decreases with the distance $l$ traveled according to the law $v = v_0 - \beta l$, where $v_0$ is the initial velocity, and $\beta$ is a constant coefficient. How does the viscous friction force acting on a body from the fluid side depend on the velocity of the body? | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| The equation of Newton's second law for the direction of motion: | |||
| $$ ma = F_с $$ | |||
| $$ \frac{F_с}{m}=\frac{dv_x}{dt} $$ | |||
| Derivative of velocity with respect to time | |||
| $$ \frac{dv_x}{dt}=\frac{d}{dt}(v_0-βx) $$ | |||
| $$ \frac{dv_x}{dt}=\frac{dv_0}{dt}-β\frac{dx}{dt}=-βt $$ | |||
| the minus sign shows that the acceleration vector is directed in the direction opposite to the velocity vector. | |||
| Combining the equations, we obtain the value of the resistance force as a function of velocity | |||
| $$ \boxed{F=βmv} $$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$F = βmv$$ | |||
| </p> | |||
| <footer class="row container"> | |||
| <br> | |||
| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small> | |||
| </p> | |||
| </footer> | |||
| </body> | |||
| </html> | |||