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PhysicsNLMMedium2 minQB

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Two blocks are in contact on a frictionless table. One has mass 'm' and the other '2m'. Same force F is applied from left and right on m and 2m. In the two cases respectively, the ratio of contact force between the blocks in the two cases will be

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Answer:

Solution:

$In both F = 3 ma a = \frac{F}{3 m}$

(1) $N_{1} = 2 ma$ (2) $N_{2} = ma$ $\frac{N _{1}}{N _{2}} = \frac{2}{1}$ [0pt] ]

Stream:JEESubject:PhysicsTopic:NLM

⏱ 2mℹ️ Source: QB

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