$2.4.20^*.$ 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?
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: \[ \boxed{x = \frac{m^2}{M^2 - m^2}L} \]
$$x = \frac{m^2}{M^2 - m^2}L$$
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