Asked by Tim
It took 1800J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.
Answers
Answered by
Henry
W = F*d = F * (5-2) = 1800 Joules.
3F = 1800
F = 600 N.
k = 600N./3m = 200N/m
3F = 1800
F = 600 N.
k = 600N./3m = 200N/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.