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III. Long Answer Questions · Q3

Q.Derive an expression for the elastic energy stored per unit volume of a wire.

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✓ Free question

Step 1. Consider a wire of un-stretched length L, cross-sectional area A, stretched (within the elastic limit, no energy lost) to a final extension l.

Step 2. The small work done in stretching the wire a further dldl, at an intermediate extension l′l', is dW=F dl′dW=F\,dl', so the total work done from 0 to l is W=∫0lF dl′W=\int_0^l F\,dl'.

Step 3. Using Young's modulus, Y=F/Al′/L⇒F=YAl′LY=\dfrac{F/A}{l'/L}\Rightarrow F=\dfrac{YAl'}{L}; substituting, W=∫0lYAl′L dl′=YAL[l′22]0l=YAl22LW=\int_0^l\dfrac{YAl'}{L}\,dl'=\dfrac{YA}{L}\left[\dfrac{l'^2}{2}\right]_0^l=\dfrac{YAl^2}{2L}.

Step 4. Rewriting, W=12(YAlL)l=12FlW=\dfrac12\left(\dfrac{YAl}{L}\right)l=\dfrac12 Fl: the elastic potential energy stored equals half the product of the final force and the final extension.

Step 5. Dividing by the wire's volume ALAL gives the energy density (energy per unit volume): u=WAL=12(FA)(lL)=12×stress×strainu=\dfrac{W}{AL}=\dfrac12\left(\dfrac{F}{A}\right)\left(\dfrac{l}{L}\right)=\dfrac12\times\text{stress}\times\text{strain}, which, using stress = Y times strain (Hooke's law), can also be written u=12Y(strain)2u=\dfrac12 Y(\text{strain})^2.

✓Final answer

The elastic energy stored per unit volume of a stretched wire is u=12×stress×strain=12Y(strain)2u=\dfrac12\times\text{stress}\times\text{strain}=\dfrac12 Y(\text{strain})^2, derived by integrating dW=F dldW=F\,dl using F=YAl/LF=YAl/L and dividing the total work W=12FlW=\tfrac12 Fl by the wire's volume AL.

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