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.

1 Answer
Apr 24, 2018

#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.