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Standard XII

Physics

Pulley

A block is ha...

Question

A block is hanging at one end of the massless rope passing over a fixed smooth pulley. From other end of the rope a man of same mass climbs with an acceleration " $x$ " relative to the acceleration of rope which is $a$ . The value of " $x$ " is :

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\frac{a}{2}

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\frac{5 a}{2}

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Solution

The correct option is D 2a
Magnitude of acceleration of string equals acceleration of block.
For man, $T - M g = M a_{m a n}$
For block, $T - M g = M a$
$∴ a = a_{m a n} .$
String accelerates downwards with a, and man acclerates upwards with a.
Hence relative acceleration $x = 2 a$

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A block is hanging at one end of the massless rope passing over a fixed smooth pulley. From other end of the rope a man of same mass climbs with an acceleration "x" relative to the acceleration of rope which is a. The value of "x" is :

The correct option is D 2aMagnitude of acceleration of string equals acceleration of block.For man, T−Mg=MamanFor block, T−Mg=Ma∴a=aman.String accelerates downwards with a, and man acclerates upwards with a.Hence relative acceleration x=2a

A block is hanging at one end of the massless rope passing over a fixed smooth pulley. From other end of the rope a man of same mass climbs with an acceleration " $x$ " relative to the acceleration of rope which is $a$ . The value of " $x$ " is :

The correct option is D 2a
Magnitude of acceleration of string equals acceleration of block.
For man, $T - M g = M a_{m a n}$
For block, $T - M g = M a$
$∴ a = a_{m a n} .$
String accelerates downwards with a, and man acclerates upwards with a.
Hence relative acceleration $x = 2 a$