Question

proove the potential energy per unit volume = ½× stress × strain
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Answer 1

Suppose F is applied on a wire of length l due to which the length of wire increased by Δl .

Before this force is applied the internal restoring force in the wire is zero.

When the length increased by Δl, the internal restoring force will be increased from 0 to F

so, the average internal force for an increase in length Δl of wire is = $\frac{0+F}{2} = \frac{F}{2}$

so work done on wire is, W = average force x increase in length

   ⇒  W = F/2 x Δl

This work done is stored as elastic potential energy U in the wire, so

     U = F/2 x Δl

 ⇒ U = stretching force x increase in length

Let A be the area of cross section of the wire, then

U = (1/2) x (F/A) x (ΔL/L) x (AL)

⇒ U = 1/2 x (stress) x (strain) x (volume of the wire)

so elastic potential energy per unit volume of the wire is given by

u = U / (volume)

so, u = 1/2 x stress x strain .

HOPE THIS HELPS YOU !!.

Answer 2

Answer:

Suppose F is applied on a wire of length l due to which the length of wire increased by Δl .

Before this force is applied the internal restoring force in the wire is zero.

When the length increased by Δl, the internal restoring force will be increased from 0 to F

so, the average internal force for an increase in length Δl of wire is = \frac{0+F}{2} = \frac{F}{2}

2

0+F

=

2

F

so work done on wire is, W = average force x increase in length

⇒ W = F/2 x Δl

This work done is stored as elastic potential energy U in the wire, so

U = F/2 x Δl

⇒ U = stretching force x increase in length

Let A be the area of cross section of the wire, then

U = (1/2) x (F/A) x (ΔL/L) x (AL)

⇒ U = 1/2 x (stress) x (strain) x (volume of the wire)

so elastic potential energy per unit volume of the wire is given by

u = U / (volume)

so, u = 1/2 x stress x strain .

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