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

PhysicsW.E.P.Conservative force & potential energyMedium2 min

PhysicsMediummultiple choice

The spring is compressed by a distance a and released. The block again comes to rest when the spring is elongated by a distance b. During this process

Options:(select one or more)

Answer:

B, C

Solution:

Work done by the spring on block = loss in spring P.E.

$= \frac{1}{2} k a^{2} - \frac{1}{2} k b^{2} = \frac{1}{2} k (a^{2} - b^{2})$

this is also work done against friction $= μ mg (a + b)$

$∴ μ = \frac{\frac{1}{2} k ( a ^{2} - b ^{2} )}{mg ( a + b )} = \frac{k}{2 mg} (a - b)$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Conservative force & potential energy

⏱ 2m

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