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Physics - NLM Question with Solution | TestHub

PhysicsNLMSimple Application of NLMMedium2 minPYQ_2012

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Three blocks of masses 4 kg. 2 kg. 1 kg respectively are in contact on a frictionless table as shown in the figure. If a force of $14 N$ is applied on the 4 kg block, the contact force between the $4 kg$ and the $2 kg$ block will be

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Answer:

Solution:

We know that, $F = ma$ $a = \frac{F}{m} = \frac{14}{7} = 2 ms^{- 2}$ Hence, from the figure $14 - N = 4 a$ $14 - N = 8 N = 6 N$

Stream:WBJEESubject:PhysicsTopic:NLMSubtopic:Simple Application of NLM

⏱ 2mℹ️ Source: PYQ_2012

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