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(C) The normal force $N$ exerted by the surface on the block is equal to its weight: $N = mg = 70 	imes 10 = 700\,N$.The maximum static friction forc

Vedclass · Accepted Answer

$700\,N \le F \le 754\,N$

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(C) The normal force $N$ exerted by the surface on the block is equal to its weight: $N = mg = 70 	imes 10 = 700\,N$.The maximum static friction force (limiting friction) is $f_{max} = \mu N = 0.4 	imes 700 = 280\,N$.Since the block is not moving,the applied horizontal force $P$ satisfies $0 \le P \le 280\,N$,and the static friction force $f$ adjusts itself such that $f = P$. Thus,$0 \le f \le 280\,N$.The net contact force $F_{net}$ exerted by the surface on the block is the vector sum of the normal force $N$ and the friction force $f$: $F_{net} = \sqrt{N^2 + f^2}$.Since $N = 700\,N$ (constant) and $0 \le f \le 280\,N$,the range of the net contact force is:Minimum value: $f = 0 \Rightarrow F_{net} = \sqrt{700^2 + 0^2} = 700\,N$.Maximum value: $f = 280\,N \Rightarrow F_{net} = \sqrt{700^2 + 280^2} = \sqrt{490000 + 78400} = \sqrt{568400} \approx 754\,N$.Therefore,$700\,N \le F \le 754\,N$.

$A$ block of mass $70\,kg$ is kept on a rough horizontal surface $(\mu = 0.4)$. $A$ person is trying to pull the block by applying a horizontal force,but the block is not moving. The net contact force exerted by the surface on the block is $F$,then

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