Question

When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is :
(1) Mgl
(2) MgL
(3) (1/2)Mgl
(4) (1/2)MgL

Answer 1

Answer:

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

The elastic potential energy stored in the extended wire is Mgl. Option first is correct.

Explanation:

• Given data

        Mass of object =M (kg)

        $\mathbf{\textrm{Gravity acceleration } =(\frac{m}{s^{2}})}$

        Initial length of wire= L (m)

        Final length of wire= L+l  (m)

• If we take final position is reference as datum position.

        Then

         $\mathbf{\textrm{Initial potential energy} \ (U_{i})=Mgl}$

• Finally when block come to datum position. It potential energy  

         become zero.

         $\mathbf{\textrm{Final potential energy} \ (U_{f})=0}$

• At final position wire is strained. In this time energy stored in wire is       strain energy.

        $\mathbf{\textrm{Strained energy} \ (E_{f})=E}$

• So by conservation of energy.

        Initial Energy = Final Energy

        $\mathbf{U_{i}=U_{f}+E_{f}}$

        $\mathbf{Mgl=0+E}$

        So E = mgl = Strain energy stored in wire

        Means option first is correct.