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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)?
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
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