14-6 Kinetics of a Particle: Work and Energy | Chapter 14: Hibbeler Dynamics | Engineers Academy

[@AdityaT03]

Great video. Thank you.

[@mikec9333]

Thank you!

[@Mr_Hilder]

comentario en esapñol porque mi profesor de dinamica trabaja con estos problemas y un gran video

[@MGRO7-d8i]

When we apply the breaks isn’t there opposite force acting with the fraction force?

[@EngineersAcademy2020]

Yes there is friction force f=UkN

[@agmrafiuzzaman8252]

Sir this math can be solved in only two lines, using v^2= u^2 + 2as formula twice. In first case we get the "a" and then using this a in 2nd case we get s=12m. No friction here, so no complexity

[@EngineersAcademy2020]

U can solve it using kinematics equation, but this problem is from chapter 14 which applies the Work Energy Principle

[@redroses4679]

Sir what is the equation used in case 2? Where is this found ?