Two blocks of mass 4kg and 2kg are placed side by side on a smooth horizontal surface (as shown in the figure). A horizontal force of 20N is applied on 4kg block. Find
i) The acceleration of each block
ii) normal reaction on the block B due to A

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$\frac{10}{3} m / s^{2}, \frac{20}{3} N$

$\frac{15}{3} m / s^{2}, \frac{25}{3} N$

$\frac{25}{3} m / s^{2}, \frac{15}{3} N$

$\frac{20}{3} m / s^{2}, \frac{10}{3} N$

answer is A.

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Detailed Solution

$m_{1} = 4 kg$

$F = 20 N$

$m_{2} = 2 kg$

$a = \frac{F}{m_{1} + m_{2}}$

$= \frac{20}{4 + 2}$

$= \frac{20}{6}$

$a = \frac{10}{3} m / s^{2}$

$F_{12} = \frac{m_{2} F}{m_{1} + m_{2}}$

$= \frac{2 \times 20}{4 + 2}$

$= \frac{40}{6} = \frac{20}{3} N$

$F_{12} = \frac{20}{3} N$

(b) Alternate method From the force body diagram The net force actually on the block B is normal force exerted by block ‘A’ only

$∴$ N on Bexerted by $A = m_{B} a$

$= 2 \times \frac{10}{3}$

$= \frac{20}{3} N$

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