A 1.0 kg block of wood sits on top of an ide | vedclass

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(D) Let $m$ be the mass of each block. The total mass of the system is $2m$.For the bottom block to move,the force of friction between the two wood bl

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$2\mu_2 < \mu_1$

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(D) Let $m$ be the mass of each block. The total mass of the system is $2m$.For the bottom block to move,the force of friction between the two wood blocks $(f_1)$ must be greater than or equal to the force of friction between the bottom block and the table $(f_2)$.The maximum static friction between the wood blocks is $f_{1,max} = \mu_1 N_1 = \mu_1 mg$.The maximum static friction between the bottom block and the table is $f_{2,max} = \mu_2 N_2 = \mu_2 (2mg) = 2\mu_2 mg$.For the bottom block to move along with the top block,the force $f_1$ must be able to overcome $f_{2,max}$.Thus,$f_{1,max} \ge f_{2,max}$.Substituting the values: $\mu_1 mg \ge 2\mu_2 mg$.Dividing both sides by $mg$,we get $\mu_1 \ge 2\mu_2$.

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