Springs and Work

2 Joules of work is needed to stretch a spring from its natural length of 30cm to a length of 42 centimeters. How much work is needed to stretch it from 35cm to 40cm?

Sorry, I know this is simple but I’m messing up somewhere because I’m not getting the correct answer of 25/24 or approx. 1.04 Joules

Any help is appreciated. Thanks. :slight_smile:

Please show your work, rather then asking other people to do the work for you.

PS. I get 1.04 joules.

I’m not trying to get anyone to do any work for me. I did a bunch of my own work already spending a lot fo time on this. However rather than posting everything that I did wrong, maybe someone would be GRACIOUS enough to post the one thing they did right.

I would like to compare someone’s right work to my wrong work to see where I went wrong and correct my error so I can understand. I don’t want you all to have to analyze my work, I just need to see how it is done properly, and then I can analyze my own work and gain an understanding myself.

It is clear I don’t know how do do this problem correctly. I think you alll know me well enough that it’s clear that I like to do my own work, not get other people to do it for me. I’m trying to gain an understanding, not find an “easy way out.”

EDIT: You know, it’s late, I’m tired, and I’m trying to find an answer than I don’t have but I know someone does. And even worse, they know I need it too. I’m not a big fan of wasting time. I don’t do this too often but I think I need to add a few of these :mad: :mad: :mad:

And then when someone is willing to actually help, maybe I’ll post a few of these :slight_smile: :slight_smile:

F = a x (where x is the “stretch” of the spring)

W is the integral of F, from 0 to the “stretch” of the spring

W = (a x^2) / 2

For a stretch of (42-30) you have that the work is 2 joules
Solve this for a, you now have the formula for work as
function of stretch of the spring. (a=4/144 in this case)

At this point, you subtract the work for a stretch of (40-30) cm
from the work for (35-30) cm and you get the difference.

In other incarnations of this problem, watch out for the
possibility that the “stretch” of the spring is actually traveling
through the natural length, where the spring is
returning energy for a portion of the travel.

Eugene

Okay, this is helpful. I never thought of this approach. I assume you meant 42 instead of 40. Anyway, how do I find the work for 30-35?

I assumed I would need to solve this problem by finding the spring constant k (you called it a, any letter will do)

When I solve for this I get 277.7777 N/m for the spring constant

when I try to find the work from 30-35, I assume we can set 30 = to 0 because it is the natural length of the spring. Then 35 would actually be 5 cm of change in length. So, if we put .05 meters into Work = (1/2)(kx^2) I get .347 N-m or Joules of Work. If I subtract this from 2 Joules of Work for an extend from 30-42 (or 0-12 depending on how you look at it) I don’t get 1.04, I get 1.65

Any ideas?

Attached are my hand calculations. I get 1.04 Joules.

-Paul

spring.pdf (27.6 KB)


spring.pdf (27.6 KB)

[quote=sanddrag]I assume you meant 42 instead of 40.
\QUOTE]

I don’t think so. The stretch from 30 to 42 centimeters
requires a work of 2 joules. Given the formula for
work, W = (a/2) stretch^2 you calculate a, or if you
are hung up on variable names, k.
The result, using the data you have been given, is 4/144.
Don’t get hung up on the units, it is okay to use joules
per cm^2 as long as you are consistent…

You are then asked for the work to stretch from 35 to 40
cm, and you can get this by subtracting the work required
to stretch from 30 to 35, W1, from the work required to stretch
from 30 to 40, W2.

W2 = (4/144) * (1/2) * (40-30)^2
W1 = (4/144) * (1/2) * (35-30)^2

W2-W1 = (2/144) * (100 - 25) = 150/144 = 25/24 joules

This is how a physicist views the world,
for how an engineer views it refer to Paul’s
correctly done calculation. Either view
can be used to solve the problem…[/quote]

from my double maths

EPE (elastic potential energy / work done from natural length) = lambda x^2 /2l
i’ve attached a spreadsheet
first, work out value of the spring force thing (grr can’t remember proper name)
even better you can work out lambda/2l
from the working you can see you get this to be 2/144
then use this value to work out the work done in extending the spring from the point with no extention, to x=5 and x=10, and work out the difference

i get 25/24, or1.041667 :slight_smile:
Hope this helps

energgy.xls (14.5 KB)


energgy.xls (14.5 KB)

you made a mistake, you forgot that is should be over 2l , not 2
:slight_smile:
it means that when rounding you got a similar answer but its not quite the same as 25/24.
:o

elastic.bmp (65.4 KB)


elastic.bmp (65.4 KB)

No. It is NOT over 2L. Natural length has nothing to do with elastic potential energy. The only thing that matters is delta x from the zero energy point, which is the natural length. Notice that your spreadsheet never actually calculates Lamda (spring constant) and you never use the 30 (natural length)in your calculation. If you did use the 30 you would get a spring constant of:

2 Joules (or N-m) = (Lamda * .12m * .12m)/(2*.3m), which yields 83.333N,

which is not the correct spring constant and not even the right units for spring constant. The correct calculation is:

2 Joules (or N-m) = (Lamda * .12m * .12m)/(2), which yields 277.778 N/m for spring constant. This yields exactly 25.24 I just rounded for simplification.

Remember, there is no work done at natural length … none, zero, nada.

Work is the integral of force per unit length (for linear motion). The force of a linear spring is defined as Kx. The integral over x of Kx is 1/2K*x^2. Remember, that in this definition x is defined as zero at the natural length.

I am 100% confident that my calculation is exactly correct given the problem’s boundary conditions.

ah, spring constant is actually the modulus of elasticity over length, and i used the modulus of elasticity… you have used the spring constant…so we are both right… :stuck_out_tongue:

Whoa there, Paul, take a deep breath :wink:

I am not 100% confident in almost anything. But that’s just me…

On the topic at hand, my first thought was “The problem doesn’t say the spring was linear.”

But, then again, my thesis was in Non-linear Vibrations, so I may just have a Jones for non-linearities…

Taking non-linear off the table, I prefer to attack these type of problems in geometric terms.

As has been pointed out, the work is the area under the force-displacement curve.

The 2 joule thing is actually the area of a triangle that is .12m wide at the base – figuring this out, I get that the triangle height is 2*2/.12= 33N.

This gives me the spring constant (33N/.12m = 280N/m )*

Given that, it is easy enough to see that area under the curve between the two points in question (.35 and .40 m) is the force of the spring at the mid-point times the base:

[the force at the middle of these two numbers F(.075m) = 21 N] X

[the distance between these two numbers (.05m)] =

1.1 J

The same result as everyone else I suppose, but perhaps just a bit different slant on things.

Joe J.

*note in my mind I have to keep straight that “m” is actually best thought of in this case as “meter (of stretch)” All units are not the same, even when they are measured with a stick with similarly spaced ticks. Most times it is fine, right and just to treat units like any other variable. But there are times when they actually do stand for different things and cancelling them can cause you grief.