FE Exam Review - Dynamics (Kinematics)

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[@saiyangodbroly26]

Much quicker way to solve for this is to use an energy balance:
Potential Initial + Kinetic Initial = Potential Final + Kinetic Final
Treat initial height as reference -> PE initial = 0.
Block comes to a rest at the top -> KE final = 0.
1/2mv^2 = mgh, m cancels.
Write h in terms of the distance up the incline plane. Sin30 = h /d -> h = d*sin30
1/2*(5)^2 = (9.81)*d*sin30 -> d = 2.55 m.

[@manjurelahi2640]

Just use V2=U2+2as, where V=0 because it will stop and you already have acceleration. So, you find the distance traveled S.

[@amanhaf]

Thank you for the tutorial!
In the section where you are showing the "Particle Rectillinear Motion" equations you will be using around 6:50 the last equation does have time at all ( vf^2=vi^2+2ai(sf-si)) which makes it faster, assuming we are not needing the time as well
vf = 0 since it stops, vi = 5 m/s, ai = -4.905 m/s^2, si = 0 because it stared from 0, and sf is what we are looking for we get an sf value of 2.55. Is this a corrert way of going about it?

[@directhubfeexam]

Thank you! Yes you can absolutely use that equation. It makes the work a lot easier too! Good catch, I’m glad you pointed that out.

[@efazchowdhury3866]

Easier way derived from 1/2*m*v^2=mgh
sqrt(2gh)=v
(v^2 /(2g))=h
plugging the numbers in gives you that h= 1/274
h is the vertical component of d, so just divide h by sin(30) giving you 2.55

[@guvencolak350]

You are pretty good
Thanks for your time
Really appreciated

[@brianwin5183]

thank you

[@directhubfeexam]

Thanks for watching!