1. The problem statement, all variables and given/known data A body of mass 5kg is suspended at the lower end of a mass-less string, which passing through a mass-less pulley and is being pulled down at the other end by a force of 500N. The force of the pull reduces at a rate of 30N per meter, through which the body is raised. What is the velocity of the body when it reaches a height 10m, starting from rest? At what height will the direction of velocity change? 2. Relevant equations a=dv/dt=v.dv/dx T-mg=ma T=f, where f = F-30y so basically, F-mg=ma 3. The attempt at a solution starting from F-mg=ma, i went to deriving: (F-30y)-mg=m(dv/dt)=m.v.dv/dy so, (F-mg)*y -15(y^2) = (m/2)v^2 => (500-5*10)*10 - 15*(100) = (5/2)v^2 => 4500 - 1500 = (5/2)v^2 => 1200 = v^2 so v = 10√12 (that's my answer) for the velocity to change direction, it must first become zero again - so putting zero as the initial and final values limits of integration, i got: (F-mg)*y -15(y^2) = 0 => [F -mg -15y]*y = 0 so y = 0 or (F-mg)/15 = 30 as y=0 is starting point, it cant be accepted as the required answer hence, at y=30, the block will change it's velocity IS WHAT I DID CORRECT?