Asked by Mackenzie
A force of 4 pounds is required to hold a spring stretched 0.1 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.6 feet beyond its natural length?
Answers
Answered by
bobpursley
work=INT force*dx from 0 to .6
= int(k x dx)= 1/2 k x^2 over limits
where k= 4lbs/.1f= 20 lbs/ft
work= 1/2 k x^2 from zero to x=.6
work= 1/2 (20lbs/ft)*.6^2 ft^2 - 0
= 36 ft-lbs work
= int(k x dx)= 1/2 k x^2 over limits
where k= 4lbs/.1f= 20 lbs/ft
work= 1/2 k x^2 from zero to x=.6
work= 1/2 (20lbs/ft)*.6^2 ft^2 - 0
= 36 ft-lbs work
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.