A 50.0-N box is sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at one instant the box is sliding the right at 1.75 m/s and that it stops in 2.25 s with constant acceleration. What magnitude fo

The acceleration of the box can be calculated using the formula: {eq}a = \frac{Delta v}{Delta t} = \frac{0 - 1.75}{2.25} = -0.78 m/s^2 {/eq} (since the box is slowing down in the positive direction) The force of friction can be calculated using the formula: {eq}F_f = ma = (50.0 N)(-0.78 m/s^2) = -39.0 N {/eq} (with the negative sign indicating that the force of friction acts in the opposite direction to the motion of the box) Therefore, the magnitude of the force of friction is {eq}| F_f | = 39.0 N \approx 38.9 N (rounded to one decimal place) {/eq} The correct answer is E. {eq}rm 38.9 N {/eq}.