First slide

Elastic Modulie

Question

A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. Then, the elastic energy stored in the wire is

Moderate

A

0. 2J
B

10 J
C

20J
D

0.1 J

Solution

The elastic energy stored in the wire is given by
$\begin{array}{l} U = \frac{1}{2} \times stress \times strain \times volume \\ = \frac{1}{2} \times \frac{F}{A} \times \frac{l}{L} \times AL = \frac{1}{2} Fl \\ = \frac{1}{2} \times 200 \times (1 \times 10^{- 3}) = 0 \cdot 1 J \end{array}$

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