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(C) The total force $F$ exerted by the surface on the object is the resultant of the normal reaction force $R$ and the frictional force $f$.Since the

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$Mg \le F \le Mg\sqrt{1 + \mu^2}$

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(C) The total force $F$ exerted by the surface on the object is the resultant of the normal reaction force $R$ and the frictional force $f$.Since the object is on a horizontal surface,the normal reaction is $R = Mg$.The total force exerted by the surface is $F = \sqrt{f^2 + R^2}$.When the object is not moving,the frictional force $f$ can range from $0$ to the limiting friction $\mu R$,i.e.,$0 \le f \le \mu Mg$.When $f = 0$,$F = \sqrt{0^2 + (Mg)^2} = Mg$.When $f = \mu Mg$ (limiting friction),$F = \sqrt{(\mu Mg)^2 + (Mg)^2} = Mg\sqrt{\mu^2 + 1}$.Therefore,the value of $F$ lies between $Mg$ and $Mg\sqrt{1 + \mu^2}$,i.e.,$Mg \le F \le Mg\sqrt{1 + \mu^2}$.

An object of mass $M$ is placed on a rough horizontal surface (coefficient of friction $\mu$). $A$ person tries to pull the object by applying a horizontal force,but the object does not move. The force $F$ exerted by the surface on the object is:

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