vikas askiitian expert
Last Activity: 13 Years ago
let mass per unit length is k ... k = M/L
now , after some time let x part of chain is above the ground
mass of part above ground = kx
let constant force applied by external source = F then
F - (kx)g = (kx)a (a is accleration)
F = (kx)g + kx(a) ....................1
from above eq , a = [F - (kx)g]/kx
a = [F/kx - g] ................2
now , a = dv/dt
multiply both sides by dx
a(dx)= dv(dx/dt) {dx/dt = v }
adx = vdv = v2/2 ............3
from eq 2 & 3
[F/kx - g ]dx = v2/2
integrating both sides
[Flnx/k - gx] = v2
v = { [Flnx/k - gx ]}1/2 ...................4