Edit to “Statement”
en/2.4.2.md
+1 −2
| @@ -1,7 +1,6 @@ | |||
| ### Statement | |||
| − | $2.4.2.$ A small body of mass | ||
| − | . The rod is given a longitudinal velocity v | ||
| + | $2.4.2.$ 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 | |||
| \[\boxed{E_2 = \frac{m_1v^2}{2}-Fl}\] | |||
| unchanged lines 16 | |||
| @@ -1,7 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $2.4.2.$ A small body of mass |
$2.4.2.$ 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 $v$ | ||
| . 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 ? | 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 | ### Solution | ||
| Initial kinetic energy of a system is: | Initial kinetic energy of a system is: | ||
| \[E_1= \frac{m_1v^2}{2}\] | \[E_1= \frac{m_1v^2}{2}\] | ||
| The difference between the initial and final kinetic energies is equal to the work done by the friction: | The difference between the initial and final kinetic energies is equal to the work done by the friction: | ||
| \[E_1-E_2 = Fl\] | \[E_1-E_2 = Fl\] | ||
| So the final kinetic energy of the system is: | So the final kinetic energy of the system is: | ||
| \[E_2 = \frac{m_1v^2}{2}-Fl\] | \[E_2 = \frac{m_1v^2}{2}-Fl\] | ||
| #### Answer | #### Answer | ||
| \[\boxed{E_2 = \frac{m_1v^2}{2}-Fl}\] | \[\boxed{E_2 = \frac{m_1v^2}{2}-Fl}\] | ||
| unchanged lines 16 | |||