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Byju's Answer

Standard XII

Physics

Pulley

In the arrang...

Question

In the arrangement shown in ﬁgure $m_{A} = m_{B} = 2 k g$ . String is massless and pulley is frictionless. Block B is resting on a smooth horizontal surface, while friction coefﬁcient between blocks A and B is $μ = 0.5$ . What maximum horizontal force F can be applied so that block A does not slip over the block B? $(g = 10 m / s^{2})$

25N

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40N

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30N

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20N

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Solution

The correct option is D
20N

Net horizontal force on block B is zero.
Hence, the given figure (a) can be replaced by figure (b).

Block B moves due to friction.
Therefore, maximum common acceleration of the two blocks can be
$a_{m a x} = \frac{f_{m a x}}{m_{B}} = \frac{10}{2} = 5 m / s^{2} ∴ F_{m a x} = (m_{A} + m_{B}) a_{m a x} = (2 + 2) (5) = 20 N$

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The correct option is D 20N Net horizontal force on block B is zero. Hence, the given figure (a) can be replaced by figure (b). Block B moves due to friction. Therefore, maximum common acceleration of the two blocks can be amax=fmaxmB=102=5 m/s2∴Fmax=(mA+mB)amax=(2+2)(5)=20N