Answer:
l =11cm,
Explanation:
let the original length of spring is l
then we can write as
[tex]3 = \int_{15-l}^{13-l} kx dx[/tex]
[tex]3 = [\frac{1}{2}kx^2]_{13-l}^{15-l}[/tex]
[tex]3 = \frac{1}{2}k[(15-l)^2 - (13-l)^2][/tex] ............1
and
[tex]5 = \int_{17-l}^{15-l} kx dx[/tex]
[tex]5 = [\frac{1}{2}kx^2]_{15-l}^{17-l}[/tex]
[tex]5 = \frac{1}{2}k[(17-l)^2 - (15-l)^2][/tex] .....................2
divide equation 2/1
[tex]\frac{5}{3} = \frac{(17-l)^2-(15-l)^2}{(15-l)^2-(13-l)^2}[/tex]
[tex]\frac{5}{3} =\frac{2(32-2l)}{2(28-2l)}[/tex]
[tex]\frac{5}{3} = \frac{16-l}{14-l}[/tex]
after solving we get
l =11cm,
The natural length of the spring is mathematically given as
[tex]L=2\sqrt{2}[/tex]
Question Parameters:
If 3 J of work are needed to stretch a spring from 13 cm to 15 cm and 5 J are needed to stretch it from 15 cm to 17 cm
Generally using the energy of spring formula we have
3=1/2K(15-L^2)-(13-L^2)
5=1/2K(17-L^2)-(15-L^2)
Therefore
3/5=(1/2K(15-L^2)-(13-L^2))/(1/2K(17-L^2)-(15-L^2))
[tex]L=2\sqrt{2}[/tex]
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