*Notes to a video lecture on http://www.unizor.com*

__Work and Elasticity__

Let's consider a slightly more complicated case of a variable force.

A good example of this is the work performed by a force to stretch or

compress a spring, in which case a force linearly depends on a length a

spring is stretched or compressed.

Consider a spring with elasticity coefficient

**that force**

*k***compresses by length**

*F***.**

*S*Consider further that a spring is compressed by the same length in equal

time interval, that is its end, where the force is applied, is

uniformly moves under the force of compression with constant speed

**.**

*V*Our task is to find the amount of work needed to compress a spring by the length

**.**

*L*In this case we can calculate the length of compression as a function of time

*S(t) = V·t*According to the Hook's Law, the force of compression is proportional to the length of compression:

*F = k·S*Since the force and the length of compression are time-dependent, this can be written as

*F(t) = k·S(t) = k·V·t*An infinitesimal increment (

*differential*) of work equals

*d*

**W(t) = F(t)·**d**S(t) = F(t)·V·**d**t =**

= k·V·t·V·d= k·V·t·V·

**t = k·V²·t·**d**t**The time interval

**needed to compress a spring by length**

*T***with speed of compression**

*L***equals to**

*V*

*T = L/V*Let's integrate the differential of work on a time interval [

**]:**

*0,T*

*∫*_{[0,T]}

*d*

**W(t) = k·V²·∫**_{[0,T]}

*t·**d*

**t =**

= k·V²·T²/2 = k·L²/2= k·V²·T²/2 = k·L²/2

Remarkably, the work does not depend on speed

**of compression, only on the length**

*V***the spring is compressed and its elasticity**

*L***.**

*k*We can easily generalize this by getting rid of dependency on the speed of compression in our calculations.

Instead of all parameters being functions of time

**, let's use the length of compression**

*t***as a base variable.**

*S*Since

**,**

*F(S) = k·S**d*

**W(S) = F(S)·**d**S = k·S·**d**S**We can integrate it on an interval

**∈[**

*S***] getting**

*0,L*

*∫*_{[0,L]}

*d*

**W(S) = k·∫**_{[0,L]}

*S·**d*

**S =**

= k·L²/2= k·L²/2

As you see, this is a more general (and simpler!) derivation of the same

formula that does not depend at all on how we compress the spring, only

on its elasticity and a length of compression.

So, total amount of work

**to compress a spring with elasticity**

*W***by length**

*k***equals**

*L*

*W = k·L²/2*regardless of how exactly we compress a spring.

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