Question bank/JEE Main/Physics

JEE MainPhysicsModerateNewton's Laws of Motion

Block A is given an acceleration 12ms−212{ms}^{- 2} towards left as
shown in Figure. Assume block B always remains horizontal, find the

acceleration
( in ms−2)\left( \text{ in }{ms}^{- 2} \right) of B.

Show answer and solution
l1+l2=C⇒l1′′+l2′′=0\mathcal{l}_{1} + \mathcal{l}_{2} = C \Rightarrow \mathcal{l}_{1}^{''} + \mathcal{l}_{2}^{''} = 0

Source: Grass-G / P-06 Physics paper 14 October.docx

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