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Let us imagine a rough horizontal surface with static frictional coefficient $u$ and an object of mass $m$ sitting on that surface. The maximum static frictional force that can be applied is $uN$ where $N$ is the normal reaction. So,if we apply a force $F<uN$,then the object won't move. However,as soon as we apply a force equal to limiting friction,the object will be on the verge of moving. Now let's apply a large force $F>uN$. What will be the acceleration of the object now? Will it be $\frac{F}{m}$ or $\frac{F-uN}{m}$? I am confused about the latter one since the friction will stop working after $F$ is applied,so shouldn't the acceleration be only $\frac{F}{m}$.

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The kinetic friction force is generally less than the static friction force, i.e., in general $\mu_{k}N\lt \mu_{s}N$. So if the applied force $F$ that causes the transition from static to kinetic friction to occur continues to be applied when sliding occurs, then the applied force is greater than the kinetic friction force and

$$a=\frac{F-u_{k}N}{m}$$

Refer to the friction plot here:http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kin

Hope this helps.

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According to the standard basic model of friction we learn in introductory physics, at the instant that is starts moving, kinetic friction takes over. The kinetic friction force always has magnitude $\mu_kN$, where $\mu_k<\mu_s$, and so the acceleration becomes $$ a=\frac{F_{\textrm{total}}}{m}=\frac{F-\mu_kN}{m}\,. $$ The statement $\mu_k<\mu_s$ that the coefficient of kinetic friction is less than the coefficient of static friction is an empirical fact that applies within the regime of applicability of this simple model of friction.

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