2.4.2 | правка #15271 | Решения Савченко

Statement

2.4.2. A small body of mass $$m_2$$ is placed in the middle of a rod of mass $$m_1$$ and length $$2l$$ . The rod is given a longitudinal velocity $$v$$
by a blow. At this time, the body will slide down the rod. What will be the kinetic energy of the system after this, if the friction force is equal to F ?

Solution

Initial kinetic energy of a system is:

$E_1= \frac{m_1v^2}{2}$

The difference between the initial and final kinetic energies is equal to the work done by the friction:

$$$E_1-E_2 = Fl$$$

So the final kinetic energy of the system is:

$E_2 = \frac{m_1v^2}{2}-Fl$

Answer

$\boxed{E_2 = \frac{m_1v^2}{2}-Fl}$