Question

The bob of a pendulum is released from a horizontal position $A$ as shown in figure. If the length of the pendulum is $2m$, what is the speed with which the bob arrives at the lowermost point $B$, give that it dissipate $10\%$ of its initial potential energy w.r.t. $B$ point against air resistance? $(g=10{m/s}^2)$

Answer 1

At position A potential energy of the particle will be , $U=mgh=20m$

Energy dissipated = $10$ %of potential energy =$\frac{10}{100}\times 20m=2m$

so, kinetic energy at point B will be potential energy -dissipated energy

$K.E=20m-2m=18m$

$\frac{1}{2}m{v}^2=18m$

$v^2=36\Rightarrow v=6m/s$