# A box of mass m is pushed horizontally on a rough floor with an initial speed of 2 m*s^(−1). The coefficient of kinetic friction between the surface and the box is `mu_k` = 0.1. Calculate the distance the box will move before stopping. The acceleration of the box can be found from the second Newton's Law:

`vecF = mveca` , where `vec F ` is the net force acting on the box. The only force on the box in the horizontal direction is the force of friction, which equals

`F= mu_k mg` (the...

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The acceleration of the box can be found from the second Newton's Law:

`vecF = mveca` , where `vec F ` is the net force acting on the box. The only force on the box in the horizontal direction is the force of friction, which equals

`F= mu_k mg` (the coefficient of kinetic friction times the normal force on the box, which in turn equals the weight of the box, because there is no other forces involved in the vertical direction).

Thus, from  `mu_k mg = ma` , the acceleration is `a = mu_kg` .

The distance d the box will move before stopping can be found from the equation of motion that involves velocity, distance and acceleration:

`v_f^2 - v_0 ^2 = -2ad`

` ` Note that acceleration is taken with the negative sign, because the frictional force opposes the direction of motion. The acceleration has the direction opposite to the direction of the initial velocity `vec v_0` . The final velocity is 0, because the box will stop.

`0 - 2^2 = -2*0.1*9.8d`

From here `d = 4/(2*0.1*9.8) = 2.04 ` meters.

The box will move a distance of 2.04 meters before stopping.

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