Question

heyaa

A wire is suspended by one end. At the other end a weight equivalent to 20 N force is applied. If the increase in length is 1.0 mm. the increase in energy of the wire will be (1) 0.013 (2) 0.02 J (3) 0.04J (4) 1.00 J
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Answer 1

Answer:

Explanation:

Elastic energy = 1/2 ×F×X  

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F=20N ,x=1mm= 10⁻³m

∴E = 1/2 × 20 ×1 × 10⁻³ =0.01

I AM NOT SURE