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Suppose that 6 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 47 cm. (a) How much work is needed to stretch the spring from 41 cm to 43 cm? (Round your answer to two decimal places.) (b) How far beyond its natural length will a force of 20 N keep the spring stretched? (Round your answer one decimal place.)
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I first thought of setting 6=\[\int\limits_{0}^{.11}kx\] and getting k eqauling 991.7355
Then I did \[\int\limits_{.05}^{.07} kx\] but that equaled 495.81, which seems kind of high
???
why do you need integral?
W=F.S We need k right F=-kx
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So we have W=-kx^2 the displacement s=x
it is better to draw the figure|dw:1454038145826:dw|
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