Properties of matterMedium
Question
A wire suspended vertically from one of its ends stretched by attaching 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
Options
A.0.2 J
B.10 J
C.20 J
D.0.1 J
Solution
The elastic potential energy stored in the wire,
U =
× stress × strain × volume
=
× Al = 
Fᐃl = × 200 × 10-3 = 0.1 J
U =
× stress × strain × volume=
× Al = Fᐃl = × 200 × 10-3 = 0.1 JCreate a free account to view solution
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