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?

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    <h3 id="back-link"><a href="/#2.4">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <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?

    <h3>Solution</h3>
    <p>
        The dog pulls the sled, performing work equal to 

$$A = \int_0^L F \, ds$$
Where:
$$F = \mu mg$$
So we get:
$$A = \mu mg L$$
According to the work-energy theorem, we know that:
$$A = \Delta K = \frac{mv^2}{2}$$
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.
$$mv = (M + m)u$$
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:
$$\frac{(M + m) u^2}{2} = \mu (M - m) g x$$
From there we can derive:
$$x = \frac{(M + m) u^2}{2 \mu g (M - m)}$$

We already know, that:
$$u = \frac{m}{M + m}v$$

Finally:
$$\boxed{x = \frac{m^2}{M^2 - m^2}L}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$x = \frac{m^2}{M^2 - m^2}L$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">   
      Nursultan Rayapov<br>
    </p>


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