Question

When a block of mass M is suspended by a long wire of length L the length of wire becomes (L+l). The elastic potential energy stored in the extended wire is

Answer 1

elastic Potential energy per unit volume is given by, U = 1/2 × stress × strain

Let volume of wire is V , cross sectional area is A and length of wire is already given i.e., L

so, elastic potential energy = 1/2 × stress × strain × V

now, stress = force applied/area

= weight of body/A

= Mg/A

and strain = change in length/original Length

= l/L

so,elastic potential energy = 1/2 × Mg/A × l/L × AL

= 1/2 × Mg × l

= 1/2 Mgl

hence, elasticpotential energy is 1/2Mgl

Answer 2

Answer:

elastic potential energy stored =1/2 F∆L

U=1/2Mg l

Mg because F =mass ×gravity

here L is neglected because the formula is ∆L so only the elongation is considered. Hence L is neglected.