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Question

(C) Let the acceleration of the system be $a$. For the $4 \ kg$ block to move with the $8 \ kg$ block without slipping,the maximum frictional force mu

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(C) Let the acceleration of the system be $a$. For the $4 \ kg$ block to move with the $8 \ kg$ block without slipping,the maximum frictional force must provide the necessary acceleration.The limiting friction on the $4 \ kg$ block is $f_{max} = \mu N = \mu mg = 0.4 	imes 4 	imes 10 = 16 \ N$.This frictional force provides the acceleration to the $4 \ kg$ block: $f_{max} = m_1 a \Rightarrow 16 = 4a \Rightarrow a = 4 \ m/s^2$.Now,considering the whole system of $(4 \ kg + 8 \ kg)$ moving with acceleration $a$ under force $F$:$F = (m_1 + m_2) a = (4 + 8) 	imes 4 = 12 	imes 4 = 48 \ N$.Thus,the maximum force $F$ is $48 \ N$.

Find the maximum force $F$ (in $N$) for which both the blocks will move with the same acceleration.

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