A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is \(\displaystyle 0.25\) and the push imparts an initial speed of \(\displaystyle 2.5 \ \text{m/s}\)?
If you push a box with a force \(\displaystyle F\) and there is friction on the floor, the sum of the forces are:
\(\displaystyle F - F_R = ma\)
If you remove your hand from the box, your force \(\displaystyle F\) is no longer acting on the box, so the forces become:
\(\displaystyle -F_R = ma\)
Or
\(\displaystyle -\mu mg = ma\)
Or
\(\displaystyle -\mu g = a \ \ \ \ \textcolor{red}{(1)}\)
where \(\displaystyle \mu\) is the coefficient of friction.
If we assume that the acceleration \(\displaystyle a\) is constant, we can write it in terms of the distance \(\displaystyle x\) from this equation:
\(\displaystyle v^2 = v^2_0 + 2ax\)
We know that the box will stop at some moment so the final velocity \(\displaystyle v = 0\).
\(\displaystyle 0 = v^2_0 + 2ax\)
This gives:
\(\displaystyle a = -\frac{v^2_0}{2x}\)
Plug this in equation \(\displaystyle \textcolor{red}{(1)}\)
\(\displaystyle -\mu g = \left(-\frac{v^2_0}{2x}\right)\)
Or
\(\displaystyle \mu g = \left(\frac{v^2_0}{2x}\right)\)
Plug in numbers.
\(\displaystyle 0.25(9.8) = \left(\frac{2.5^2}{2x}\right)\)
This gives:
\(\displaystyle x = \textcolor{blue}{1.28 \ \text{m}}\)
