Physics doubt 2

see here as weight is added at a rate dmdt
there is a thrust force generated in the reverse direction
so F_net = F - F_thrust
since this force reduces velocity
so the same force wuld be required to keep the belt moving at 2 m/s

so force required is F_thrust = vdmdt = 2 x 0.5 = 1 N

@manmay

wat actually is the thrust force??...n how is it generated backward when the body is falling from top??....is it another name for pseudo force??

yes the same doubt for me too
I can't understand ur explanation

Problems related to variable mass can be solved following these steps :
1. Make a list of all forces acting on the main mass and apply them on it.
2. Aplly an additional thrust force \vec{F_{t}} on the mass, the magnitude of which is \left|\vec{v_{r}} \left(\pm \frac{dm}{dt} \right)\right| where v_r is relative velocity i.e, velocity of mass gained or masses ejected relative to the main mass, and the direction of \vec{v_{r}} is given by the direction of \vec{v_{r}} in case of mass increasing and otherwise the direction of -\vec{v_{r}} if mass is decreasing .
3. Find net force on the mass and apply \vec{F_{net}} =m\frac{d\vec{v}}{dt} ( m = mass at particular instant )
4. Integrate it with proper limits to find velocity at any instant of time t

now in abv problem mass is increasing so direction is \vec{v_{r}}
hence F_net = F - F_thrust
F_net = F - \vec{F_{net}} =\vec{v_{r}}\frac{dm}{dt}

had it been decreasing
F_net = F + \vec{F_{net}} =\vec{v_{r}}\frac{dm}{dt}

oh k...understood...btw how does ur post get pink??