Sol:

Given:

Mass, \(M_{1} \) = 100kg

Mass, \(M_{2} \) = 200kg

g = \(10m/s^{2} \)

The coefficient of friction between a and b,\(\mu _{1} = 0.2 \)

The coefficient of friction between b and the ground, \(\mu _{2} = 0.3. \)

Frictional force acting between blocks A and B = \(f_{1} = \mu _{1}M_{1}g \) \(\Rightarrow \) 0.2 * 100 * 10

\(\Rightarrow \)200 N

Frictional force acting between blocks A and B = \(f_{2} = \mu _{2}(M_{1} + M_{2})g \) \(\Rightarrow \)0.3 * (100 + 200) * 10

\(\Rightarrow \)0.3 * (300) * 10

\(\Rightarrow \) 900 N

The minimum force required to move block B = \(f = f_{1} + f_{2} \) \(\Rightarrow \)200 N + 900 N = 1100 N

Therefore, the minimum force required to move block B is 1100 N.

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