On a rough horizontal surface a body of mass 2 kg is given a velocity of 10 m/s. If the coefficient of friction is 0.2 and `g=10 m/s^2,` the body will stop after covering a distance of _m?

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Denote the mass of a body as `m` and the coefficient of (kinetic) friction as `mu.`

There are three forces acting on this body: the gravitational force `mg` downwards, the reaction force `N` upwards and the friction force `F_f` which acts at a direction opposite to the movement.

The movement is horizontal, therefore vertical forces are balanced, `N=mg.` Also it is known that `F_f=mu N=mu mg.` By Newton's Second law the magnitude of the acceleration a is `F/m=mu g.`

Denote the initial speed as `V_0.` The acceleration `a` has the opposite direction. For a body moving with a constant acceleration the speed `V=V_0-at` and the displacement `D=V_0t-(at^2)/2.`

A body stops when `V=0,` i.e. at `t=V_0/a,` and the displacement is

`V_0^2/a-V_0^2/(2a)=V_0^2/(2a)=V_0^2/(2mu g).`

In numbers, the distance will be `10^2/(2*0.2*10)` =25 (m). This is the answer.

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