Asked 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.)
Answered by 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

Answered by CALCULUS IS KILLING ME
YO YOU'RE WRONG
Answered by baxter
do Damons thing and divide by 2 at the end.
Answered by Hey
he is not off on his equations. I followed Damon's instructions to a similar equation and now I understand what I am doing! Thank you for your help!
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