Physics, asked by supersayian, 9 months ago

a spring of constant 100N/m is stretched upto 5 cm.find out workdone

Answers

Answered by JunaidMirza

3

Answer:

0.125 J

Explanation:

Work done on a spring is given as

Work = 0.5kx²

= 0.5 × 100 N/m × (0.05 m)²

= 0.125 J

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