Question

A particle moves on a rough horizontal ground with some initial velocity say v₀. If $\frac{3}{4}$ of its kinetic energy is lost due to friction in time t₀ then coefficient of friction between the particle and the ground is:-
(1) $\frac{{\mathrm{v}}_0}{2{\mathrm{gt}}_0}$ (2) $\frac{{\displaystyle {\mathrm{v}}_0}}{{\displaystyle 4{\mathrm{gt}}_0}}$ (3) $\frac{{\displaystyle 3{\mathrm{v}}_0}}{{\displaystyle 4{\mathrm{gt}}_0}}$ (4) $\frac{{\displaystyle {\mathrm{v}}_0}}{{\displaystyle {\mathrm{gt}}_0}}$

Answer 1

$$\begin{gathered} initialKE \\ K{E}_i=\frac{m{v_0}^2}{2} \\ energylossinfrictionistheworkdonebythefriction. \\ Workdonebyfriction \\ W=\left(\frac{3}{4}\right)\frac{m{v_0}^2}{2}=\frac{3m{v_0}^2}{8} \\ speedbev \\ \frac{m{v}^2}{2}=\frac{m{v_0}^2}{8} \\ v=\frac{v_0}{2} \\ letthecoefficientbe\mu \\ a=-\mu g \\ -\mu g=\frac{\frac{v_0}{2}-v_0}{t_0} \\ \mu g=\frac{\frac{v_0}{2}}{t_0} \\ \mu =\frac{v_0}{2g{t}_0} \\ REgards \end{gathered}$$