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A spring of force constant 10N / m has a...

A spring of force constant 10N / m has an initial stretch 0.20 m . In changing the stretch to 0.25 m , the increase in potential energy is about

A

0.1 joule

B

0.2 joule

C

0.3 joule

D

0.5 joule

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta P.E. = (1)/(2)k(x_(2)^(2)-x_(1)^(2))=(1)/(2)xx10[(0.25)^(2)-(0.20)^(2)]`
`= 5xx0.45xx0.05=0.1 J`
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