The potential energy in an elastic string of force constant K which has been extended by X metres is expresses as
a
\(\frac{1}{2}k \Box ^2\)
b
\(k \Box ^2\)
c
\(\frac{1}{2}k \Box \)
d
\(k \Box \)
Explanation
Correct Option
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sulaimanharuna
3 years ago
P.E =work done in the elastic string =average force ×distance(extension)
0+f/2×x
=1/2fx
But, from hooke's law I.e f=kx
.: 1/2kx² option A is the correct answer (jamb 2022 past question)
