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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
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| <p class="author"> | | <p class="author"> |
| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
| </p> | | </p> |
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| <h3 id="back-link"><a href="../../#2.4">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.4">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.4.20^*.$ A dog of mass m is tied with a leash of length $L$ to a sledge of mass $M > m$. At the initial moment, the dog is next to the sled. What is the greatest distance a dog can move the sledge in one jerk, if the coefficients of friction of the dog's paws and the sled runners on the horizontal surface are the same? | | $2.4.20^*.$ A dog of mass m is tied with a leash of length $L$ to a sledge of mass $M > m$. At the initial moment, the dog is next to the sled. What is the greatest distance a dog can move the sledge in one jerk, if the coefficients of friction of the dog's paws and the sled runners on the horizontal surface are the same? |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="180" /> | | loading="lazy" width="180" /> |
| <figcaption> | | <figcaption> |
| For problem $2.4.20^*$ | | For problem $2.4.20^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| The dog pulls the sled, performing work equal to | | The dog pulls the sled, performing work equal to |
| \[ | | \[ |
| A = \int_0^L F \, ds | | A = \int_0^L F \, ds |
| \] | | \] |
| Where: | | Where: |
| \[ | | \[ |
| F = \mu mg | | F = \mu mg |
| \] | | \] |
| So we get: | | So we get: |
| \[ | | \[ |
| A = \mu mg L | | A = \mu mg L |
| \] | | \] |
| According to the work-energy theorem, we know that: | | According to the work-energy theorem, we know that: |
| \[ | | \[ |
| A = \Delta K = \frac{mv^2}{2} | | A = \Delta K = \frac{mv^2}{2} |
| \] | | \] |
| After the dog has run a distance $L$, the leash tethered to the sled will stretch. This scenario can be analyzed as an inelastic collision. | | After the dog has run a distance $L$, the leash tethered to the sled will stretch. This scenario can be analyzed as an inelastic collision. |
| \[ | | \[ |
| mv = (M + m)u | | mv = (M + m)u |
| \] | | \] |
| In this scenario, the work done against the frictional force represents the energy lost by the system, which results in the deceleration of the sled-dog system. This loss of kinetic energy can be quantified using the work-energy principle: | | In this scenario, the work done against the frictional force represents the energy lost by the system, which results in the deceleration of the sled-dog system. This loss of kinetic energy can be quantified using the work-energy principle: |
| \[ | | \[ |
| \frac{(M + m) u^2}{2} = \mu (M - m) g x | | \frac{(M + m) u^2}{2} = \mu (M - m) g x |
| \] | | \] |
| From there we can derive: | | From there we can derive: |
| \[ | | \[ |
| x = \frac{(M + m) u^2}{2 \mu g (M - m)} | | x = \frac{(M + m) u^2}{2 \mu g (M - m)} |
| \] | | \] |
| | | |
| We already know, that: | | We already know, that: |
| \[ | | \[ |
| u = \frac{m}{M + m}v | | u = \frac{m}{M + m}v |
| \] | | \] |
| | | |
| Finally: | | Finally: |
| \[ | | \[ |
| \boxed{x = \frac{m^2}{M^2 - m^2}L} | | \boxed{x = \frac{m^2}{M^2 - m^2}L} |
| \] | | \] |
| | | |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
