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## Homework Statement

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?

## Homework Equations

a=dv/dt=v.dv/dx

T-mg=ma

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:

(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 can't be accepted as the required answer

hence, at

**y=30**, the block will change it's velocity

**IS WHAT I DID CORRECT?**