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Three blocks of masses $5 k g$ , $3 k g$ and $2 k g$ are placed on rough surface $(μ_{k} = 0.1)$ as shown in figure. A horizontal force of $20 N$ is applied on $5 k g$ block. The force exerted by $3 k g$ block on $2 k g$ block is (in Newton):

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answer is 4.

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Detailed Solution

If $a$ is the acceleration of the blocks, then

$20 - μ N = 10 a$

or $20 - 0.1 \times 10 g = 10 a$

$∴ a = 1 m / s^{2}$

Now $F_{1} - f = 2 a$

or $F_{1} - 0.1 \times 2 g = 2 \times 1$

$∴ F_{1} = 4 N$

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