The frictional force acting on the 1 ,kg bloc | vedclass

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(A) Assume both blocks move together with a common acceleration $a$. The total mass of the system is $M = 1 \,kg + 100 \,kg = 101 \,kg$. The external

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$0.1$

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(A) Assume both blocks move together with a common acceleration $a$. The total mass of the system is $M = 1 \,kg + 100 \,kg = 101 \,kg$. The external force applied is $F = 10 \,N$. Using Newton's second law,$F = Ma$,we get $10 = 101 	imes a$,so $a = \frac{10}{101} \approx 0.099 \,m/s^2 \approx 0.1 \,m/s^2$. The frictional force $f$ acting on the $1 \,kg$ block provides the acceleration to it. Thus,$f = m 	imes a = 1 \,kg 	imes 0.1 \,m/s^2 = 0.1 \,N$. The maximum limiting friction is $f_L = \mu N = 0.5 	imes 1 \,kg 	imes 10 \,m/s^2 = 5 \,N$. Since the required frictional force $(0.1 \,N)$ is less than the limiting friction $(5 \,N)$,the blocks will move together,and the frictional force acting on the $1 \,kg$ block is $0.1 \,N$.

The frictional force acting on the $1 \,kg$ block is .................. $N$.

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