# Q18  The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

Consider the extreme position (horizintal) :-

The kinetic energy at this position is zero as velocity is zero.

Thus total energy is given by :    $=\ mgl\ +\ 0 \ =\ mgl$

Now consider the mean position (lowermost point) :

Here the potential energy of bob is zero.

Whereas   kinetic energy is  :

$=\ \frac{1}{2}mv^2$

Further, it is given that 5 per cent of energy is dissipated due to air resistance while coming down.

Thus energy equation becomes (conservation of energy):-

$\frac{1}{2}mv^2\ =\ \frac{95}{100}\times mgl$

or

$v\ =\ \sqrt{\frac{2\times 95\times1.5\times 9.8}{100}}\ =\ 5.28\ m/s$

Exams
Articles
Questions