1 Work Power Energy
A spring of force constant 10 N per meter has an initial stretch 0.20 m. In changing the stretch to 0.25m, the increase in potential energy is about 1) 0.1 joule 2) 0.2 joule 3) 0.3 joule 4) 0.5 joule
SOLUTION:−
Elastic potential energy of spring = ½ kx2 Change in energy = ½10 ( 0.252 - 0.202) Change in energy = 0.1125 J
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