A spring requires a force of 25 N to produce an extension of 2 cm in it. What is the work done in further extending it by 3 cm.

Question

A spring requires a force of 25 N to produce an extension of 2 cm in it. What is the work done in further extending it by 3 cm.

Answer 1

Hi Haley!

 

It turns out that, due in large part to the spring force being conservative, the amount of work done to stretch or compress a spring is equal to the change in the spring's potential energy caused by the extension or compression.  So:

 

W = ΔU_s

 

where W is the work done ON the spring, and ΔU_s is the change in the spring's potential energy as a result of that work.  Basically, in this case, it means that if you do a certain amount of work in stretching the spring, you store an equal amount of potential energy in the spring.

 

[I will state that without demonstrating it, but we can demonstrate it, if you like]

 

So, the question here is, what is the change in potential energy of the spring between being stretched from 2 cm of extension to 5 cm of extension?  For a spring, the expression for potential energy is:

 

U_s = (1/2)kx²

 

where x is the amount of extension and k is the spring constant (or force constant).  The spring constant is just a measure of how much force it takes to extend or compress the spring by a certain length (stiffer springs have higher force constants because it takes more force to stretch them by the same amount as a "springier" spring).  Within the range where they act as springs, springs follow Hooke's law where:

 

F = -kx  (the negative sign just means the spring pushes back against compression and pulls back against extension)

 

So, as a matter of magnitude, k = F/x -- and, as long as the spring is following Hooke's law, the ratio of F/x will be a constant of the spring (the spring would no longer follow Hooke's law only if you, say, extended it so far that you deformed the spring; in other words, passed the "elastic limit" where it would spring back)

 

So we can calculate k by knowing it takes 25 N to extend the spring 2 cm:

 

k = (25 N)/(0.02 m) = 1250 N/m  (I changed the length unit to meters, to make sure we get joules as the unit for energy and work later on)

 

Knowing this, we can now calculate the spring potential energy the spring possesses when extended to 2 cm and 5 cm.

 

U_s2cm = (1/2)(1250 N/m)(0.02 m)² = 0.25 J  at 2 cm extension

U_s5cm = (1/2)(1250 N/m)(0.05 m)² = 1.56 J at 5 cm extension

 

Note that the spring has increased in potential energy (as you may expect) during its extension.  The amount by which the energy increased between those two extensions is equal to the amount of work done on the spring to extend it from 2 to 5 cm.

 

I hope this helps!  If you have any more questions about this or similar problems, just let me know.