Question

A body of mass 25kg moving at 3ms-1 on rough horizontal floor is brought to rest after sliding through a distance of 2.5m on the floor. Calculate the friction of sliding friction(take g=10ms-2)

Answer 1

Let's first find the deceleration of the body:

$$v^2=v_0^2+2ad,$$ $$a=\dfrac{v^2-v_0^2}{2d},$$ $$a=\dfrac{0-(3\ \dfrac{m}{s})^2}{2\cdot2.5\ m}=-1.8\ \dfrac{m}{s^2}.$$

The sign minus means that the body decelerates.

Let's apply the Newton's Second Law of Motion:

$$-F_{fr}=ma,$$ $$-\mu_smg=ma,$$ $$\mu_s=-\dfrac{a}{g}=-\dfrac{-1.8\ \dfrac{m}{s^2}}{10\ \dfrac{m}{s^2}}=0.18$$