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?
T=f, where f = F-30y
so basically, F-mg=ma
The Attempt at a Solution
starting from F-mg=ma, i went to deriving:
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 can't be accepted as the required answer
hence, at y=30, the block will change it's velocity
IS WHAT I DID CORRECT?