If the mass of block A = 10,kg,the coefficien | vedclass

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(A) To start the motion,the tension $T$ in the string must overcome the limiting static friction force $f_{s,max}$.For block $A$ on the horizontal sur

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$2\,kg$

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(A) To start the motion,the tension $T$ in the string must overcome the limiting static friction force $f_{s,max}$.For block $A$ on the horizontal surface,the normal force $N = m_A g$.The limiting static friction is $f_{s,max} = \mu_s N = \mu_s m_A g$.For block $B$ hanging vertically,the tension $T = m_B g$.At the point of starting motion,the system is in limiting equilibrium,so $T = f_{s,max}$.Substituting the values: $m_B g = \mu_s m_A g$.$m_B = \mu_s m_A = 0.2 	imes 10\,kg = 2\,kg$.Therefore,the mass of block $B$ required to start the motion is $2\,kg$.

If the mass of block $A = 10\,kg$,the coefficient of static friction $\mu_s = 0.2$,and the coefficient of kinetic friction $\mu_k = 0.2$,find the mass of block $B$ required to start the motion.

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