Question

a box is is given a push so that it slides across the floor. How far will it go, given that the coefficient for kinetic friction is 0.15 and the push imparts an initial speed of 3.5m/s?

Answer 1

Let's apply the work-kinetic energy theorem:

$$\Delta KE=W,$$ $$KE_f-KE_i=-F_{k.fr}d,$$ $$0-\dfrac{1}{2}mv^2=-\mu_k mgd,$$ $$d=\dfrac{1}{2}\dfrac{v^2}{\mu_kg},$$ $$d=\dfrac{1}{2}\cdot\dfrac{(3.5\ \dfrac{m}{s})^2}{0.15\cdot 9.8\ \dfrac{m}{s^2}}=4.2\ m.$$

Answer:

$d=4.2\ m.$