Question

when a box of mass 4.5kg is given in an initial speed of 5m/s, it slides along horizontal flooor to a distance of 3m before coming to rest. what is the co-effiicient of the kinetic friction between the box and the floor?

Answer 1

Let's first find the deceleration of the box:

$$v^2=v_0^2+2ad,$$ $$0=v_0^2+2ad,$$ $$a=-\dfrac{v_0^2}{2d}=-\dfrac{(5\ \dfrac{m}{s})^2}{2\times3\ m}=-4.16\ \dfrac{m}{s^2}.$$

Applying the Newtons Second Law of Motion, we get:

$$-F_{fr}=ma,$$ $$-\mu mg=ma,$$ $$\mu=-\dfrac{a}{g}=-\dfrac{-4.16\ \dfrac{m}{s^2}}{10\ \dfrac{m}{s^2}}=0.416$$