Asked by Ray
Six and one-half foot-pounds of work is required to compress a spring 4 inches from its natural length. Find the work required to compress the spring an additional one-half inch. (Round your answer to two decimal places.)
This is what I did so far:
4 in = 1/3 ft
6.5 = ∫[0,1/3] kx dx
6.5 = k(x^2/2)(Eval at 1/3 and 0)
6.5 = k/18
117 = k
Now figuring the Work:
W = ∫[1/3,5/6] 117x dx
W = 117(x^2/2) (Eval at 5/6 and 1/3)
W = 34.125 ft-lb (34.13 ft-lb) but Webassign is counting that wrong. I re-looked at my problem and everything but I can't seem to find what's wrong with my work. Help please?
This is what I did so far:
4 in = 1/3 ft
6.5 = ∫[0,1/3] kx dx
6.5 = k(x^2/2)(Eval at 1/3 and 0)
6.5 = k/18
117 = k
Now figuring the Work:
W = ∫[1/3,5/6] 117x dx
W = 117(x^2/2) (Eval at 5/6 and 1/3)
W = 34.125 ft-lb (34.13 ft-lb) but Webassign is counting that wrong. I re-looked at my problem and everything but I can't seem to find what's wrong with my work. Help please?
Answers
Answered by
Ray
Woops, I didn't convert 1/2 inch to feet. the W equation should be [1/3,9/24], leading to the answer to be around 1/73 ft-lb
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