Question

A body moving along the ground with a velocity of $14m/s$ comes to rest due to friction after travelling a distance of $50cm$ Determine the coefficient of friction between the body and the ground $g=9.8m/s^2$

Answer 1

By applying the law of motion

$v^2=u^2+2as$

${\left(14\right)}^2=-2\times 0.5\times a$

$a=-196m/s^2$

The coefficient of friction is given as

$\mu =\frac{ma}{mg}$

The coefficient of friction is$20$