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Q.A spring of spring constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is

(a) 16 J
(b) 8 J
(c) 32 J
(d) 24 J
Jharkhand JacJAC Intermediate Board (1st Year) 2026MCQ· 1mImportance★★★★★
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Work done in stretching a spring FROM one extension TO another (not from zero) is W = (1/2) k (x2^2 - x1^2), since spring PE = (1/2)kx^2 at each extension.

Spring potential energy at extension x: U(x) = (1/2) k x^2

Work done in extending from x1 = 5 cm = 0.05 m to x2 = 15 cm = 0.15 m equals the change in spring PE:

W = U(x2) - U(x1) = (1/2) k (x2^2 - x1^2)

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