Physics+ Newton Laws, Problem 3: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 3

Problem

A chain of infinitesimally small links with overall length L and mass M is hung vertically by its top end A.
Its bottom end B touches a flat platform with a spring under it.
Assume that a platform and a spring are weightless, so the spring is in neutral position.
The elasticity coefficient of a spring is k.
Gravity acceleration is g.


If the chain is let go from point A, it will fall down on a platform, which will squeeze the spring.
What is the maximum distance the platform will go down, shortening the spring?

Solution

We will compare the potential energy of a chain in its initial position to potential energy of a squeezed spring.
As the chain falls down to a platform, its potential energy is gradually transformed into kinetic energy, which, in turn, is transformed into an energy of a spring.

When chain is fully on a platform, the energy of an oscillating spring (kinetic plus potential) should be equal to an initial potential energy of a chain.
At that time a spring will oscillate between maximum squeeze and maximum elongation with its total energy at the end points be only potential and equal to initial potential energy of a chain before it started to fall down.

Let's take the level of a platform in its neutral position as level zero of potential energy.
The mass density of a chain per unit of length is
μ = M/L
An infinitesimal piece of a chain of length dy has mass μ·dy.
Its potential energy on a height y above the surface of a platform in its neutral position, which is level zero, is
dP = μ·g·dy·y = (M/L)·g·y·dy
Potential energy of the whole chain is
P = ∫_[0,L](M/L)·g·y·dy =
= (M/L)·g·y²/2|_[0,L] = M·g·L/2
Check units:
(kg/m)·(m/sec²)·m² =
= (kg·m/sec²)·m = watt

According to the Hook's Law, the force of resistance of a spring F is proportional to the length of its shortening x.
F = k·x
When a spring is squeezed from the length x to x−dx, the work performed by an external force that caused it is
dW = F·dx = k·x·dx
To shorten a spring by the length S we have to do the work
W = ∫_[0,S]k·x·dx = k· S²/2
Check units:
[(kg·m/sec²)/m]·m² =
= (kg·m/sec²)·m = watt

From the Law of Energy Conservation follows that the potential energy of a chain in its initial position should be equal to the work performed by the force of a falling chain to squeeze a spring under a platform, that is
P = W
from which we derive
M·g·L/2 = k·S²/2
and
S = √M·g·L/k

Answer:
A spring will shorten its length by
S = √M·g·L/k
Check units:
√kg·(m/sec²)·m/(kg/sec²) = m