Question

If it requires 5.0 J of work to stretch a particular spring by 2.2 cm from its equilibrium length, how much more work will be required to stretch it an additional 3.6 cm ?

Answer 1

`29.75J`

Step-by-step explanation:

Fist we must find the constant of the spring. The equation for work in a spring is:

`W=(1)/(2)kx^2`

where
`W`, is the work, in this case
`W=5J`,
`k` is the constant of the spring, and
`x` is the distance from the equilibrium length:
`x=2.2cm=0.022m`

clearing for
`k` and substituting known values:

`2W=kx^2\\(2W)/(x^2)=k`

`(2(5J))/((0.022m)^2)=(10J)/(4.84x10^(-4)m^2) =20,661.16J/m^2`

So work to stretch the spring, an additional 3.6cm is:

`W=(1)/(2)kx^2`

where
`x=2.2cm+3.6cm=5.8cm=0.058m`

`W=(1)/(2)(20,661.16J/m^2)(0.058m)^2`

`W=(1)/(2)(20,661.16J/m^2)(3.364x10^(-3)m^2)`

`W=(1)/(2)(69.5J)`

`W=34.75J`

And since we already have 5J, the difference, or the additional work is:

`34.75J-5J=29.75J`