Block A is in equilibrium. Now block A given velocity V = g k m in downward direction. Maximum normal reaction on block B is :-

By W.E.T. − 2 1 k ( x 1 2 − x 1 2 ) + mgx = 0 − 2 1 mv 2 max . comp x = k 2 mg kx = 2 mg

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

PhysicsMediumsingle choice

Block A is in equilibrium. Now block A given velocity $V = g \frac{m}{k}$ in downward direction. Maximum normal reaction on block B is :-

Answer:

Solution:

By W.E.T.

$- \frac{1}{2} k (x_{1}^{2} - x_{1}^{2}) + mgx = 0 - \frac{1}{2} mv^{2} max . comp x = \frac{2 mg}{k} kx = 2 mg$

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

⏱ 2m

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