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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.MiscellaneousEasy2 minPYQ_2014

PhysicsEasysingle choice

A spring of spring constant $5 \times 1 0^{3} N m^{- 1}$ is stretched initially by $5 cm$ from the unstretched position. The work required to stretch it further by another $5 cm$ is

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Answer:

Solution:

$U_{1} = \frac{1}{2} k x_{1}^{2} = \frac{1}{2} \times (5 \times 1 0^{3}) \times (5 \times 1 0^{- 2})^{2}$ $= 6.25 N m$ $U_{2} = \frac{1}{2} k x_{2}^{2} = \frac{1}{2} \times 5 \times 1 0^{3} \times (5 + 5)^{2} \times 1 0^{- 4}$ $= 25 N m$ $∴$ Work done $= U_{2} - U_{1} = 25.0 - 6.25$ $= 18.75 N m$

Stream:AIIMSSubject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2014

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