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(B) Given: Mass of the block $m = 5 \ kg$,applied horizontal force $F = 1000 \ N$,and coefficient of friction $\mu = 0.1$. The acceleration due to gra

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$50$

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(B) Given: Mass of the block $m = 5 \ kg$,applied horizontal force $F = 1000 \ N$,and coefficient of friction $\mu = 0.1$. The acceleration due to gravity is taken as $g = 10 \ m/s^2$.$1$. The normal force $N$ exerted by the wall on the block is equal to the applied horizontal force: $N = F = 1000 \ N$.$2$. The limiting friction $f_L$ is given by $f_L = \mu N = 0.1 	imes 1000 = 100 \ N$.$3$. The weight of the block acting downwards is $W = mg = 5 	imes 10 = 50 \ N$.$4$. For the block to be in equilibrium,the upward frictional force $f$ must balance the downward weight $W$. Thus,the required frictional force is $f_{req} = W = 50 \ N$.$5$. Since the required frictional force $(50 \ N)$ is less than the limiting friction $(100 \ N)$,the block remains in equilibrium,and the actual frictional force is $50 \ N$.

For the given figure,if the block remains in equilibrium,find the frictional force between the block and the wall in $N$.

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