Calculus
posted by Kaitlyn
Suppose that 5 J of work is needed to stretch a spring from its natural length of 28 cm to a length of 36 cm.
(a) How much work is needed to stretch the spring from 32 cm to 34 cm? (Round your answer to two decimal places.)
(b) How far beyond its natural length will a force of 15 N keep the spring stretched? (Round your answer one decimal place.)

Damon
W = (1/2) k x^2
where x is the stretch or compress length
here x = .36  .28 = .08 meter
so
5 = .5 k (.0064)
k = 1563 N/m so (1/2) k = 781
a)
work to .34  work to .32
subtract the .28 from both
.06 and .04
= 781 (.06^2.04^2)
=1.562 J
b) F = k x
15 = 1563 * x
x = .00960 m
= .96 cm 
CALCULUS IS KILLING ME
YO YOU'RE WRONG
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