Правка разделов «Statement», «Solution», «Answer»
en/2.3.9.md
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| ### Statement | |||
| − | $2.3.9.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
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| $2.3.9.$ A slingshot was made from a long rubber band of stiffness $k$. Find the kinetic energy of a "projectile" fired from this slingshot if the rubber band was stretched with a force $F$ and then released. | |||
| ### Solution | |||
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| When the slingshot is stretched by a distance $x$ from both ends, the total pulling force $F$ is balanced by the tension in the two halves of the rubber band. Let's write the equilibrium condition: | |||
| $$\frac{F}{2} = kx$$ | |||
| Expressing the elongation $x$ of the rubber band: | |||
| $$x = \frac{F}{2k} \quad (a)$$ | |||
| The potential energy of the stretched rubber band at this point is: | |||
| $$E_{\text{pot}} = \frac{kx^2}{2}$$ | |||
| Substituting expression (a) into the energy formula: | |||
| $$E_{\text{pot}} = \frac{kF^2}{8k^2} = \frac{F^2}{8k}$$ | |||
| From the law of conservation of energy, all the potential energy of the elastic band $E_{\text{pot}}$ converts into the kinetic energy of the projectile $E_k$: | |||
| $$E_{\text{pot}} = E_k \implies E_k = \frac{F^2}{8k}$$ | |||
| @@ -25,7 +21,3 @@Solution | |||
| #### Answer | |||
| $$E_k = \frac{F^2}{8k}$$ | |||
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| − | #### Answer | ||
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| − | [Insert a concise answer or boxed result] | ||
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| ### Statement | ### Statement | ||
| $2.3.9.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.3.9.$ A slingshot was made from a long rubber band of stiffness $k$. Find the kinetic energy of a "projectile" fired from this slingshot if the rubber band was stretched with a force $F$ and then released. | $2.3.9.$ A slingshot was made from a long rubber band of stiffness $k$. Find the kinetic energy of a "projectile" fired from this slingshot if the rubber band was stretched with a force $F$ and then released. | ||
| ### Solution | ### Solution | ||
|  | |||
| When the slingshot is stretched by a distance $x$ from both ends, the total pulling force $F$ is balanced by the tension in the two halves of the rubber band. Let's write the equilibrium condition: | When the slingshot is stretched by a distance $x$ from both ends, the total pulling force $F$ is balanced by the tension in the two halves of the rubber band. Let's write the equilibrium condition: | ||
| $$\frac{F}{2} = kx$$ | $$\frac{F}{2} = kx$$ | ||
| Expressing the elongation $x$ of the rubber band: | Expressing the elongation $x$ of the rubber band: | ||
| $$x = \frac{F}{2k} \quad (a)$$ | $$x = \frac{F}{2k} \quad (a)$$ | ||
| The potential energy of the stretched rubber band at this point is: | The potential energy of the stretched rubber band at this point is: | ||
| $$E_{\text{pot}} = \frac{kx^2}{2}$$ | $$E_{\text{pot}} = \frac{kx^2}{2}$$ | ||
| Substituting expression (a) into the energy formula: | Substituting expression (a) into the energy formula: | ||
| $$E_{\text{pot}} = \frac{kF^2}{8k^2} = \frac{F^2}{8k}$$ | $$E_{\text{pot}} = \frac{kF^2}{8k^2} = \frac{F^2}{8k}$$ | ||
| From the law of conservation of energy, all the potential energy of the elastic band $E_{\text{pot}}$ converts into the kinetic energy of the projectile $E_k$: | From the law of conservation of energy, all the potential energy of the elastic band $E_{\text{pot}}$ converts into the kinetic energy of the projectile $E_k$: | ||
| $$E_{\text{pot}} = E_k \implies E_k = \frac{F^2}{8k}$$ | $$E_{\text{pot}} = E_k \implies E_k = \frac{F^2}{8k}$$ | ||
| @@ -25,7 +21,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $$E_k = \frac{F^2}{8k}$$ | $$E_k = \frac{F^2}{8k}$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||