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:

Easy

A

0.2 J
B

10 J
C

20 J
D

0.1 J

Solution

Stored energy $= \frac{1}{2} \times F \times ∆ l = \frac{1}{2} \times 200 \times 10^{- 3} = 0.1 J$

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