A ball rolled on ice with a velocity 14m/s comes to rest after travelling 40m. Find the coefficient of friction.

Here, $u=14 m/s, v=0, s= 40 m$
Let, $g=9.8 m/s^2$
Using third equation of motion, we get
                                                         $v^2=u^2+2as$
or,                                                    
                                                       $$\begin{gathered} v^2-u^2=2as \\ 0-(14{)}^2=2\times a\times 40 \\ So, a=-\frac{14\times 14}{80}=-2.45 \end{gathered}$$
Now, using the formula of coefficient of friction, we get
                                                                                 $\mu =\frac{a}{g}=\frac{2.45}{9.8}=0.25$