Help with calculus 2!!! its difficult..10 points?
A force of 3 pounds is required to hold a spring stretched 0.5 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 1 feet beyond its natural length?
can anyone solve that???
- hace 1 décadaRespuesta preferida
For every 3 pounds you stretch the spring 1/2 feet beyond its natural length which could only mean that its spring constant is 6 pounds per feet. Work is defined as force acting through a distance, so you could say that the change of work over a distance is kx dx or 6x dx. The work required to stretch the spring 1 feet could be found from evaluating the integral of dW = 6x dx from 0 to 1. The answer should be 3 ft-lbs.
- cermakLv 4hace 4 años
no longer particularly! I in basic terms had a try on countless sequences or series immediately. it isn't any longer difficult there is in basic terms a lot of attempt to bear in mind the thank you to tutor a chain is convergent or divergent. it is extremely consumer-friendly stuff at an engineering math direction point. i do no longer understand how series is for finished blown math significant scholars yet for engineering scholars it isn't any longer that onerous. some questions are rather difficult yet no longer something you would be unable to be sure in an afternoon or 2. ;D you will do high-quality if your smart. Cal a million replaced into childs play comapred to cal 2. Cal1 isn't calculus it is in basic terms fucntions and relatives.