A block of mass m=2 kg is initially at rest | vedclass

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(A) $1$. First,we determine the normal force $F_N$ acting on the block. The forces in the vertical direction are the upward force $F_2$,the normal for

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$2.5$

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(A) $1$. First,we determine the normal force $F_N$ acting on the block. The forces in the vertical direction are the upward force $F_2$,the normal force $F_N$,and the downward gravitational force $mg$. Since the block is not moving vertically,the net vertical force is zero:$F_2 + F_N - mg = 0$$10 \ N + F_N - (2 \ kg)(10 \ m/s^2) = 0$$10 \ N + F_N - 20 \ N = 0$$F_N = 10 \ N$$2$. Next,we calculate the limiting static friction force $f_{s,max}$:$f_{s,max} = \mu_s F_N = 0.4 	imes 10 \ N = 4.0 \ N$$3$. The applied horizontal force is $F_1 = 6 \ N$. Since the applied horizontal force $F_1$ is greater than the limiting static friction force $f_{s,max}$ $(6 \ N > 4.0 \ N)$,the block will start to move.$4$. Once the block is in motion,the frictional force acting on it is the kinetic friction force $f_k$:$f_k = \mu_k F_N = 0.25 	imes 10 \ N = 2.5 \ N$Therefore,the magnitude of the frictional force acting on the block is $2.5 \ N$.

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