Formula for potential energy store din a strained solid?

AI Thread Summary
The discussion revolves around finding the formula for potential energy stored in a strained solid, specifically focusing on a material called abductin used by scallops. The user presents a problem involving the elastic modulus, thickness, cross-sectional area, and compression of the material to calculate the potential energy. They initially struggle with the calculations but ultimately discover that a computational error was the cause of their incorrect answer. The correct formula for elastic potential energy is confirmed as E = λ(Ax²)/(2L). The thread concludes with the user successfully resolving their issue.
ninjagowoowoo
Messages
75
Reaction score
0
formula for potential energy stored in a strained solid??

I was wondering of there was some formula for finding the potential energy stored in a strained/stressed solid. I have this problem:

A scallop forces open its shell with a material called abductin, the elastic modulus of which is about 2.20×106 N/m2. If this piece of abductin is 2.88 mm thick and has a cross-sectional area of 0.515 cm2, how much potential energy does it store when compressed 1.19 mm?

Unfortunately my prof. didnt mention a damned thing about potential energy, so i was wondering if someone could help me out with this... oh yeah and it's not in our book either...
 
Last edited:
Physics news on Phys.org
F = \lambda \frac {Ax}{L} = T

F = force, \lambda= modulus of elasticity, A = cross sectional area x = displacement L = natural length T = tension

E = \int F dx = \lambda \frac {Ax^2}{2L}

E = elastic potential energy
 
Wow thanks a bunch. However I'm still not getting the correct answer...

I'm calculating this:


2.20x10^6 * (0.0000515 *(0.00119^2) / 0.00288*2) = 23.4075J which is incorrect.

I got it... turned out to be some weird computational error. Anyway, thanks again !
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top