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Standard XII

Physics

Newton's Laws for System of Particles

In the shown ...

Question

In the shown arrangement mass of A = 1 kg, mass of B = 2 kg. Coefficient of friction between A and B = 0.2. The re is no friction between B and the ground. The frictional force exerted by A on B equals.

2 N

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3 N

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4 N

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5 N

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Solution

The correct option is A 2 N
If there is no friction between B and ground, it will immediately move.
$a_{B} = \frac{10}{m_{A} + m_{B}} = \frac{10}{1 + 2} = \frac{10}{3} ㎨ {(f_{s})}_{m a x} = μ \times 1 \times 10 = 0.2 \times 1 \times 10 2 N F = \frac{10}{3} N > {(f_{s})}_{m a x} B l o c k A w i l l m o v e o n b l o c k B a n d f r i c t i o n a l f o r c e = {(f_{s})}_{m a x} = (f_{k i n e t i c}) = 2 N$

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In the shown arrangement mass of A = 1 kg, mass of B = 2 kg. Coefficient of friction between A and B = 0.2. The re is no friction between B and the ground. The frictional force exerted by A on B equals.

The correct option is A 2 NIf there is no friction between B and ground, it will immediately move.aB=10mA+mB=101+2=103㎨(fs)max=μ×1×10=0.2×1×102NF=103N>(fs)maxBlockAwillmoveonblockBandfrictionalforce=(fs)max=(fkinetic)=2N