## tacotime one year ago Friction Question: someone pushes a box with a mass of 11.2 kg with a constant speed of 3.5m/s. The kinetic friction is .2. If the person stops pushing how far does the box slide before coming to rest?

when it hs stopped, all the kinetic energy of the box will have been lost by friction to heat etc ie the work done by friction in retarding the box will equal the box's initial KE you can say $$\frac{1}{2}mv^2 = F_f.x = \mu mgx$$ or $$x = \dfrac{\frac{1}{2}mv^2 }{\mu mg}= \frac{v^2}{2\mu g}$$ the equations of motion will give the same result as they are essentially the same thing if you start with $$v^2 = u^2 + 2ax$$ where final velocity v = 0 you have $$x = -\frac{u^2}{2a}$$ from Newton's Law $$F = ma = -\mu mg \implies a = - \mu g$$ so $$x = \frac{u^2}{2\mu g}$$ - same result except we used u rather than v for the box's initial velocity