For the given figure,what will be the contact | vedclass

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(B) $1$. First,calculate the total mass of the system: $M_{total} = 5 \ kg + 6 \ kg + 4 \ kg + 5 \ kg = 20 \ kg$.$2$. Calculate the net force acting o

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$46$

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(B) $1$. First,calculate the total mass of the system: $M_{total} = 5 \ kg + 6 \ kg + 4 \ kg + 5 \ kg = 20 \ kg$.$2$. Calculate the net force acting on the system: $F_{net} = 90 \ N - 10 \ N = 80 \ N$.$3$. Calculate the acceleration of the system: $a = F_{net} / M_{total} = 80 \ N / 20 \ kg = 4 \ m/s^2$.$4$. To find the contact force between the $6 \ kg$ and $4 \ kg$ blocks,consider the system of the $4 \ kg$ and $5 \ kg$ blocks on the right side. Let $F_{contact}$ be the force exerted by the $6 \ kg$ block on the $4 \ kg$ block.$5$. Applying Newton's second law to the $4 \ kg$ and $5 \ kg$ blocks (total mass $9 \ kg$): $F_{contact} - 10 \ N = (4 \ kg + 5 \ kg) 	imes a$.$6$. $F_{contact} - 10 \ N = 9 \ kg 	imes 4 \ m/s^2 = 36 \ N$.$7$. $F_{contact} = 36 \ N + 10 \ N = 46 \ N$.

For the given figure,what will be the contact force applied by the $6 \ kg$ block on the $4 \ kg$ block? ............ $N$

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