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(B) The block is pressed against a vertical wall with a horizontal force $F = 1000 \, N$. The mass of the block is $m = 5 \, kg$ and the coefficient o

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$50$

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(B) The block is pressed against a vertical wall with a horizontal force $F = 1000 \, N$. The mass of the block is $m = 5 \, kg$ and the coefficient of static friction is $\mu = 0.1$.First,we calculate the maximum possible static frictional force $(f_{max})$:$f_{max} = \mu N = 0.1 	imes 1000 = 100 \, N$.Next,we calculate the downward gravitational force acting on the block:$W = mg = 5 	imes 10 = 50 \, N$ (taking $g = 10 \, m/s^2$).Since the block is in equilibrium,the upward frictional force $(f)$ must balance the downward gravitational force $(W)$.Therefore,$f = W = 50 \, N$.Since $f = 50 \, N$ is less than $f_{max} = 100 \, N$,the block remains in equilibrium,and the frictional force is $50 \, N$.

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

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