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Question: A spring of force constant 10 N/m has an initial stretch 0.20 m. In changing the stretch to 0.25 m, ...

A spring of force constant 10 N/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

Answer

0.1 joule

Explanation

Solution

∆P.E.=12k(x22−x12)=12×10[(0.25)2−(0.20)2]= \frac { 1 } { 2 } k \left( x _ { 2 } ^ { 2 } - x _ { 1 } ^ { 2 } \right) = \frac { 1 } { 2 } \times 10 \left[ ( 0.25 ) ^ { 2 } - ( 0.20 ) ^ { 2 } \right]