Asked by Anonymous
It takes 2.5 J of energy to stretch a helical spring by 15 cm. Determine the spring
constant.
constant.
Answers
Answered by
Elena
PE=kx²/2
k=2•PE/ x²=
=2•2.5/0.15²=222 N/m
k=2•PE/ x²=
=2•2.5/0.15²=222 N/m
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.