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NEET UG PHYSICS MCQ: WORK-ENERGY-POWER

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Application +4 | -1

A spring with spring constant k is extended from

x = 0\;\;\;\mathrm{to\;\;}x = {x_1}

.The work done will be

View answer and explanation

Correct option: B

\frac{1}{2}kx_1^2

Option B (

\frac{1}{2}kx_1^2

) is correct because the work done in extending a spring is equal to the change in its potential energy, which is given by the formula

\frac{1}{2}kx^2

.

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