On a smooth horizontal surface a block $A$ of mass $10 k g$ is kept. On this block a second block $B$ of mass $5 k g$ is kept. The coefficient of friction between the two blocks is $0.4$ . A horizontal force of $30 N$ is applied on the lower block as shown. The force of friction between the blocks is (take $g = 10 m / s^{2})$ .

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Solution

Ref. image I
$30 - f_{s} = 30 a$ __(i)
Ref. image II
$f_{s} = N μ$
$= (5 g) (0.4)$
$= 20$ Newtons

So $20 = 5 a_{B}$
$\Rightarrow a_{B} = 4 m / s^{2}$
& friction $= 20$ Newtons between blocks

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Ref. image I30−fs=30a __(i)Ref. image IIfs=Nμ=(5g)(0.4)=20 NewtonsSo 20=5aB⇒aB=4m/s2& friction =20 Newtons between blocks

Ref. image I
$30 - f_{s} = 30 a$ __(i)
Ref. image II
$f_{s} = N μ$
$= (5 g) (0.4)$
$= 20$ Newtons

So $20 = 5 a_{B}$
$\Rightarrow a_{B} = 4 m / s^{2}$
& friction $= 20$ Newtons between blocks