SOLUTION: the length a spring is stretched from its natural length from work. W foot-pounds, given by L = squared sign (2W/k) where k is a constant for a given spring. If a certain spring ha

Algebra ->  Length-and-distance -> SOLUTION: the length a spring is stretched from its natural length from work. W foot-pounds, given by L = squared sign (2W/k) where k is a constant for a given spring. If a certain spring ha      Log On


   

Question 159013: the length a spring is stretched from its natural length from work. W foot-pounds, given by L = squared sign (2W/k) where k is a constant for a given spring. If a certain spring has a constant of 76.4, and the spring is stretched 3.6 feet from its natural length, how much work will be necessary?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
Do you mean?
L=sqrt(2W/k)
3.6=sqrt(2W/76.4)
12.96=2W/76.4
2W=12.96*76.4
2W=990.144
W=990.144/2
W=495.072 FOOT POUNDS NEEDED.
OR?
L=(2W/K)^2
3.6=(2W/76.4)^2
3.6=(4W^2/5,836.96)
4W^2=3.6*5,836.96
4W^2=21,013.056
W^2=21,013.056/4
W^2=5,253.264
W=SQRT5,253.264
72.48 FOOT POUNDS NEEDED.