While slippling a on rough spherical surface of radius ′ R ′ , block A of mass ′ m ′ comes with velocity 1.4 g R at bottom B . Work done in slippling the block from ′ B ′ to ′ C ′ is

Loss in P . E . = gain in K . E . + w o r k done against friction m g R = 1 2 m 1.4 g R + W f W f = 0.3 m g R Now, W B → c = m g R + 0.3 m g R = 1.3 m g R

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PhysicsW.E.P.MiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

While slippling a on rough spherical surface of radius $' R'$ , block $A$ of mass $' m'$ comes with velocity $\sqrt{1.4 g R}$ at bottom $B$ . Work done in slippling the block from $' B'$ to $' C'$ is

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Answer:

Solution:

Loss in $P . E . =$

gain in $K . E . + w o r k$ done against friction $m g R = \frac{1}{2} m (1.4 g R) + W_{f}$

$W_{f} = 0.3 m g R$

Now,

$W_{B \to c} = m g R + 0.3 m g R$

$= 1.3 m g R$

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

⏱ 2mℹ️ Source: PYQ_2020

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