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a heap of chain is lying on a horizontal table a small part of the chain is released through the hole in the table . Calculate the velocity of the chain as a function of length of the vertically hanging portion

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We don't need to account for the mass

S: Distance

t: time

a : Acceleration

g : 9.81

$S(t) = \int \int {a} dt dt$

$S(t) = \frac{1}{2}a t^2$

The constants of integration are zero since the object starts at rest.

We could just use $g$ for the acceleration or we could include some drag constant $k$

$S(t) = \frac{1}{2} (g - k) t^2$

$t = {(\frac{2S}{a})}^{0.5}$

$V(t) = gt$

$V = g{(\frac{2S}{a})}^{0.5}$

or

$V = (g - k){(\frac{2S}{a})}^{0.5}$ if you prefer to include a drag constant

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  • $\begingroup$ The acceleration of the chain will vary as a function of the force acting on it due to its own weight ( which also keeps changing ) . i.e the hanging part. One can't simply take it as g let alone putting a cherry on top like (g-k) $\endgroup$
    – user32457
    Commented May 24, 2020 at 13:06
  • $\begingroup$ the force acting on it is g, which does not change unless you move to a different planet $\endgroup$
    – sav
    Commented Jun 1, 2020 at 7:53
  • $\begingroup$ g is called acceleration due to gravity or gravitational field intensity if you can comprehend . Force on anything by the virtue of it is called gravitational force : which is mg. $\endgroup$
    – user32457
    Commented Jun 4, 2020 at 8:55

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