First slide

Stress and strain

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

Easy

A

$\frac{1}{2} MgL$
B

Mgl
C

MgL
D

$\frac{1}{2} Mgl$

Solution

$Stress = \frac{F}{A} = \frac{Mg}{A}, Strain = \frac{ΔL}{L} = \frac{L + l - L}{L} = \frac{l}{L} Energy stored in the wire is, U = \frac{1}{2} \times Stress \times Strain \times Volume = \frac{1}{2} \times \frac{Mg}{A} \times \frac{l}{L} \times A \times L = \frac{1}{2} Mgl$

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