Work of 5 Joules is done in stretching a spring from its natural length to 13 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (13 cm)?(adsbygoogle = window.adsbygoogle || []).push({});

here's what i done:

W = F * D

5 = F * 13

F = 5/13

F = kx

5/13 = k(13)

k = 5/39

[tex] \int_{0}^{13} 5/39 *x dx [/tex]

i'm really confuse about this problem

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# Homework Help: Another spring question

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