How Much Extra Work Is Needed to Stretch a Spring Further?

[miamirulz29]

Homework Statement

It takes 3.23 J of work to stretch a Hooke's-law spring 12.1 cm from its unstressed length. How much extra work is required to stretch it an additional 5.76cm. Answer in Joules.

Homework Equations

F= -kx
W= FD

The Attempt at a Solution

At first, I try using proportions to do this problem:

3.23/12.1 = x/5.76
I got that answer but it was wrong.

So I tries finding the force using w=fd. Then I multiplied that by 5.76 and I got the same answer as doing it the proportion method.

What am I doing wrong. Do I need K or do I need to use the equation W = 1/2kx^2? Thanks in advance.

 

[miamirulz29]

Oh figured it out, nvm.

 

[kerimek]

I would like to clarify that the formula F=-kx is not Hooke's Law, but rather the equation for the force exerted by a spring based on its displacement from equilibrium. Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from equilibrium. This can be represented mathematically as F=-kx, where k is the spring constant.

In this problem, we are given the work (W) required to stretch the spring by 12.1 cm, which can be calculated using the formula W=FD. We can rearrange this formula to solve for the force (F) exerted by the spring, which is equal to 3.23 J/12.1 cm = 0.267 J/cm.

To find the additional work required to stretch the spring by 5.76 cm, we can use the same formula, W=FD, but this time we need to use the new force value (0.267 J/cm) and the new displacement (5.76 cm). This gives us a total work of 0.267 J/cm * 5.76 cm = 1.536 J.

Therefore, the extra work required to stretch the spring an additional 5.76 cm is 1.536 J. It is important to note that this calculation assumes that the spring constant remains constant throughout the stretching process. If the spring is stretched beyond its elastic limit, the spring constant may change and the calculation may no longer be accurate. Additionally, using the formula W= 1/2kx^2 would give the same result, as it is equivalent to the formula W=FD when the force is constant.