Physics - W.E.P. Question with Solution | TestHub

At the point$\mathrm{A}$the energy of the pendulum is entirely potential energy. At point B, the energy of the pendulum is entirely kinetic energy. It means that as the bob pendulum lowers from A to B, the potential energy is converted into kinetic energy. Thus, at$\mathrm{B}$, KE = PE. But$5\%$of the potential energy is dissipated against air resistance.



$\mathrm{KE}$at$\mathrm{B}=95\% \mathrm{of}\mathrm{ }\mathrm{PE}\mathrm{ }\mathrm{at}\mathrm{ }\mathrm{A}$.... (i)

$\therefore$From equation (i),$\frac{1}{2}m{v}^2=\frac{95}{100}mgh$

$v^2=2\times \frac{95}{100}gh=2\times \frac{95}{100}\times 9.8\times 1.5$

$v=\sqrt{\frac{19\times 9.8\times 1.5}{10}}=\sqrt{27.93}=5.285\mathrm{ }\mathrm{m} {\mathrm{s}}^{-1}$

$v=5.3\mathrm{ }\mathrm{m} {\mathrm{s}}^{-1}$