A horizontal force 12 , N pushes a block weig | vedclass

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Question

(D) The normal force $N$ exerted by the wall on the block is equal to the applied horizontal force: $N = 12 \, N$.The maximum static frictional force

Vedclass · Accepted Answer

Block will not move and force exerted on the block by the wall is $13 \, N$

Vedclass · Answer

(D) The normal force $N$ exerted by the wall on the block is equal to the applied horizontal force: $N = 12 \, N$.The maximum static frictional force $f_{s,max}$ is given by $f_{s,max} = \mu_s N = 0.5 	imes 12 \, N = 6 \, N$.The weight of the block is $W = mg = (1/2) \, kg 	imes 10 \, m/s^2 = 5 \, N$.Since the weight $W = 5 \, N$ is less than the maximum static friction $f_{s,max} = 6 \, N$,the block will not move.The wall exerts two forces on the block: the normal force $N = 12 \, N$ (horizontal) and the static frictional force $f_s = 5 \, N$ (vertical,balancing the weight).The total force exerted by the wall on the block is the resultant of these two perpendicular forces: $F_{total} = \sqrt{N^2 + f_s^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \, N$.

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