Student Question

# A block of mass `m` moves on a horizontal surface with coefficient of friction `mu` , subject to a constant force `F_0` acting vertically downward towards the surface. If the block has initial velocity `v` , how far does it travel before it stops?

There are 4 forces acting on the block:
gravitational force `mg` downwards,
the force `F_0` downwards,
the reaction force `N` upwards,
the friction force `F_f` horizontally.

The net force `F_(n et),` which is their vector sum, gives the block some acceleration `a.` By Newton's Second law `F_(n et) = ma.` Because the block moves horizontally, its speed and acceleration are also horizontal. Thus the net force is also horizontal that means all vertical forces are balanced, `N = mg + F_0,` and `F_(n et) = F_f.`

It is known that `F_f = mu N = mu(mg + F_0).` Therefore, the acceleration magnitude is `a =mu(mg + F_0)/m =mu(g + F_0/m).` Also it is known that friction force is opposite to the direction of movement.

Therefore, the speed of the block is `V(t) = v - at` and the block stops at time `t_1 = v/a` (when the speed becomes zero). The distance as a function of time is therefore  `D(t) = vt - (at^2)/2, tlt=t_1,`  and the farthest distance is

`D(t_1) = vt_1 - (at_1^2)/2 = vt_1 - (vt_1)/2 = (vt_1)/2 =v^2/(2mu(g + F_0/m)).`

This is the answer.

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