Question

prove that the elastic potential energy density of a stretched wire is equal to half the product of stress and strain?

Answer 1

Explanation:LET'....F=streching force within elastic limit

L=original length of wire ..........l=increase in length of wire,,,,,,,,,,A=area of cross section of wire ,,,,,,,,,,,,,,,,Y=young's modules of elasticity                                                                      NOW.....Y=stress/strain,,,,,,,,,,,Y=F.L/Al,,,,,,,,,,,F=YAl/L                                                                              workdone on wire by force if wire is further extended by small length [dl]             dW=F.dl,,,,,,,dW=YAl/L.dl                                                                                                                      so net workdone to extend ot from ZERO to l is,,,,,,,,,,                                                                        INT.[dW]=INT.0 TO L[YAl/L.dl]                                                       INT=intigration                                                                                                                                                                   W=1/2.Fl                                                                                                                                                                                                   this amount of workdone is stored as elastic potential energy..U in volume V                                                                                                                                                                                     Then elastic potentil stored per unit volume OR energy density=U/V=1/2.[F/A][L/l]                                                                                                                                                        =1/2.STRESS.STRAIN          ,,,,,,,,PROVED,,,,,,,,,,

Answer 2

Answer:

Elastic Potential energy (EPE) = $\frac{1}{2}$x stress x strain

Explanation:

Let's say that by using a force F, the length of the wire is raised by l.

Average internal force, = $\frac{(0 + F)}{2} = \frac{F}{2}$

Work done = $\frac{F}{2}$×$l$

The work done is stored as potential energy ⇒ Elastic Potential energy(EPE)

$= \frac{Fl}{2} = \frac{1}{2} (\frac{F}{A} )(\frac{l}{L} ).(A.L)$

where A = cross-sectional area

L = length

E.P.E = $\frac{1}2}$ x stress x strain x volume of wire

Energy density = {E.P.E./Volume of the wire} = $\frac{1}{2}$ x stress x strain

Hence Proved.

Explain the stress-strain curve.

https://brainly.in/question/17456401#:~:text=Expert%2Dverified%20answer,-question&text=To%20obtain%20the%20stress%20strain,the%20material%20follows%20Hooke's%20law.

Define stress and strain and their different types.

https://brainly.in/question/1799529

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