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Standard XII

Physics

Problem Set

The accelerat...

Question

The acceleration (in $m / s^{2}$ ) of movable pulley P is and block B is , if acceleration of block $A = 1 {m/s}^{2}$ downwards.

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Solution

From the figure, we can say:
$S M + O N + R B = c o n s t a n t$
If block B is assumed to move up, pulley P will move down so that the string between them remains taught.
$∴ v_{P} + v_{P} + (- v_{B}) = 0$
$\Rightarrow 2 v_{P} = v_{B}$
$\Rightarrow 2 a_{P} = a_{B}$

Also,
$A C + D E + F N = c o n s t a n t$
$v_{A} + 0 + (- v_{P}) = 0$
$\Rightarrow a_{A} = a_{P}$
$\Rightarrow a_{P} = 1 m / s^{2}$
$\Rightarrow a_{B} = 2 m / s^{2}$

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The acceleration (in m/s2) of movable pulley P is and block B is , if acceleration of block A=1 m/s2 downwards.

From the figure, we can say: SM+ON+RB=constant If block B is assumed to move up, pulley P will move down so that the string between them remains taught. ∴vP+vP+(−vB)=0 ⇒2vP=vB ⇒2aP=aB Also, AC+DE+FN=constant vA+0+(−vP)=0 ⇒aA=aP ⇒aP=1 m/s2 ⇒aB=2 m/s2

The acceleration (in $m / s^{2}$ ) of movable pulley P is and block B is , if acceleration of block $A = 1 {m/s}^{2}$ downwards.