Edits to “Statement”, “Solution”, “Answer”
en/2.4.2.md
+13 −3
| @@ -1,13 +1,23 @@ | |||
| ### Statement | |||
| − | $2.4.2.$ | ||
| + | $2.4.2.$ A small body of mass m2 is placed in the middle of a rod of mass m1 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 | |||
| − | [Insert a concise answer or boxed result] | ||
| + | \[\boxed{E_2 = \frac{m_1v^2}{2}-Fl}\] | ||
| @@ -1,13 +1,23 @@ | |||
| ### Statement | ### Statement | ||
| $2.4.2.$ |
$2.4.2.$ A small body of mass m2 is placed in the middle of a rod of mass m1 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 | ### 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 | #### Answer | ||
| [Insert a concise answer or boxed result] | \[\boxed{E_2 = \frac{m_1v^2}{2}-Fl}\] | ||