Question

helpme please: A spring of 3 fts requires a force of 10 pounds to stretch it to a length of 3.5 fts a) Find the work required to stretch the spring from its natural length to a length of 5 fts?

A spring of 3 fts requires a force of 10 pounds to stretch it to a length of 3.5 fts
a) Find the work required to stretch the spring from its natural length to a length of 5 fts
b) Find the work required to stretch the spring from 4 to 5 fts
c) As soon as you increase the natural length by applying a force of 30 lbs.

Answer 1

`160ft1lbs`

Explanation:

In answering this question I have looked it up and I quote from a book by S.G. Page [Mathematics, a second start] and will reference it in quotation marks where it is directly quoted.

"Hooke's law states that the tension in a an elastic spring or string is proportional to the extension"

" That is; `T=[gammax]/[L]`" where `T`=tension, `gamma`= modulus of elsticity, `L`= natural or original length and `x` = extension.

"Now the work done in stretching the spring a small distance `dx` is approximately `Tdx`..........`[1]`[ force times distance]"

" Therefore total work done=`int _0^aTdx`=`int_0^a[gammax]/Ldx` "[ from.....`1`]

So after integrating total work done = `[x^2gamma]/[2L]` from `0` to `a`

which equals, `[gammaa^2]/[2L]`, where `a` =distance stretched from original length and `gamma and L` are constants.

Young's modulus of elasticity is a constant for any given material and in this case there is enough information to calculate it directly, since we are given a force of `10`1lbs i.e,

`10`=`[gamma[0.5]^2]/[2.3]` [since in this case `a =0.5ft` and `L=3ft`],

So from his , `gamma=240` and finally the work done by stretching spring `2ft`[ i.e, from `3ft to 5ft` will equal.

`240[2^2]/2.3`=`160` ft 1lbs. [This can be converted to Si. units of energy, i.e. Joules from a conversion chart]

Hope this was helpful and not too confusing.