Given:
Work done = 87 Joule
[tex] x_{1} [/tex] = 1.4 m
[tex] x_{2} [/tex] = 2.9 m
To find:
Spring constant = ?
Formula used:
W = [tex] \frac{1}{2} k \triangle x^{2} [/tex]
Solution:
Work done required to stretch a spring from 1.4 m to 2.9 m is 87 Joule. Work done is given by following formula.
W = [tex] \frac{1}{2} k \triangle x^{2} [/tex]
where W = work done
k = spring constant or force constant
[tex] \triangle x [/tex] is change in length i. e [tex] \triangle x=x_{2} x_{1} [/tex]
k = [tex] \frac{2W}{\triangle x^{2}} [/tex]
k=[tex] \frac{87 \times 2}{[x_{2} - x_{1} ]^{2}} [/tex]
k = 77.33 N/m
Thus, spring constant of the given spring is 77.33 N/m.
