How Much Work to Stretch a Bungee Cord 1 Meter?

[Amber_]

Homework Statement

This is the problem; it's multiple choice.

When you try to stretch a bungee cord a
distance x, it resists with an opposing force of
the form b x2, where b is a constant.
If b is measured to be 6 N/m2, how much
work does it take to stretch the bungee cord a
distance of 1 meter?
1. 0.5 Joule
2. 1 Joule
3. 3 Joules
4. 2 Joules

2. The attempt at a solution

I originally thought this wasn't a hard question if the distance x is one, and b is 6 N/m^2 than the bungee cord is resisting with a force of six Newtons, meaning you need to apply at least 6 N of force over one meter (6 Joule of energy/work). However, none of the possible answers are six. I don't see where my logic is wrong here. If anyone could enlighten me, that'd be great.

Thanks!

 

[theJorge551]

The constant does not define precisely how much work is needed. The equation you're working on is F=bx where F is force, b is your constant, and x is the displacement assumed. To find the work done, you have to apply a formula, PE = 1/2 bx^2, where PE = potential energy (work, in this scenario). You can get this by integrating with respect to x, and you will get the right answer.

 

[cepheid]

Hi Amber, Welcome to PF!

To do this question properly, you have to integrate the force over the distance over which it is applied:

W = \int F(x) dx​

This is necessary because the force is not constant, but rather it varies with the displacement of the cord. In other words, the force is a function of x: F(x) = bx².

 

[cepheid]

The constant does not define precisely how much work is needed. The equation you're working on is F=bx where F is force, b is your constant, and x is the displacement assumed. To find the work done, you have to apply a formula, PE = 1/2 bx^2, where PE = potential energy (work, in this scenario). You can get this by integrating with respect to x, and you will get the right answer.

Be careful. The original problem is not actually a Hooke's Law question because the force varies quadratically with the distance, rather than linearly, and therefore the formula you gave for the elastic potential energy is not correct. The original poster will actually have to carry out the integration herself.

 

[Amber_]

The constant does not define precisely how much work is needed. The equation you're working on is F=bx where F is force, b is your constant, and x is the displacement assumed. To find the work done, you have to apply a formula, PE = 1/2 bx^2, where PE = potential energy (work, in this scenario). You can get this by integrating with respect to x, and you will get the right answer.

I thought that might be the case, and so I answered 3 joules, and apparently that is not right.

The course I'm taking uses this program that does problem sets online. You enter your answers and it tells you right away if you're right or wrong. You get so many tries at the right answer and each time you answer wrongly you get points off.

 

[PhanthomJay]

Work is force times distance only if the force is constant. In this problem, it is not constant. F = bx^2. Are you familiar with calculus and the definition of work?

Edit: beaten to the punch several times already!

 

[Amber_]

Okay, thanks!
I got it now. You're right. It's not actually a Hooke's law question. You have to integrate 6x^2, get 2x^3 and therefore the answer is 2 joules, not three.