A block can slide on a smooth inclined plane | vedclass

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(A) In a stationary lift,the acceleration of a block sliding down a smooth inclined plane is $a_{rel} = g \sin 	heta$.When the lift is moving with an

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$(g + a)\sin 	heta$

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(A) In a stationary lift,the acceleration of a block sliding down a smooth inclined plane is $a_{rel} = g \sin 	heta$.When the lift is moving with an acceleration $A$ (upward being positive),the effective acceleration due to gravity becomes $g_{eff} = g + A$.In this problem,the lift is descending with a retardation $a$. Retardation while descending means the acceleration vector is directed upwards. Thus,$A = +a$.The effective acceleration becomes $g_{eff} = g + a$.Therefore,the acceleration of the block relative to the incline is $a_{rel} = g_{eff} \sin 	heta = (g + a) \sin 	heta$.

$A$ block can slide on a smooth inclined plane of inclination $\theta$ kept on the floor of a lift. When the lift is descending with a retardation $a$,the acceleration of the block relative to the incline is

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