Question
A spring has a force constant of 546.5 N/m.
Find the potential energy stored in the
spring when the spring is
a) stretched 4.18 cm from equilibrium.
Answer in units of J
Find the potential energy stored in the
spring when the spring is
a) stretched 4.18 cm from equilibrium.
Answer in units of J
Answers
(1/2)k x^2 = (1/2)(546.5)(.0418)^2
= 0.477 Joules
= 0.477 Joules
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