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

PhysicsW.E.P.MiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A spring of spring constant $5 \times 10^{3} N m^{- 1}$ is stretched initially by $5 cm$ from the unstretched position. Then the work required to stretch it further by another $5 cm$ is

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Answer:

Solution:

$W_{1} = \frac{1}{2} k x_{1}^{2} = \frac{1 \times 5}{2} \times 10^{3} \times {(5 \times 10^{- 2})}^{2} = 6.25 J$

$W_{2} = \frac{1}{2} k {(x_{1} + x_{2})}^{2}$

$= \frac{1}{2} \times 5 \times 10^{3} {(5 \times 10^{- 2} + 5 \times 10^{- 2})}^{2} = 25 J$

Net work done $= W_{2} - W_{1}$

$= 25 - 6.25 = 18.75 J = 18.75 N m$

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

⏱ 2mℹ️ Source: PYQ_2020

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