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:

• A. 0.1 J
• B. 0.2 J
• C. 10 J
• D. 20 J

Answer 1

Answer: (A)

0.1 J

$energy=\frac{1}{2}Fl$
$=\frac{1}{2}\times 200\times \frac{1}{1000}$
$=0.1J$ .