Question

The bob of a pendulum is released from a horizontal position as shown in the figure. The length of the pendulum is $2m$. If $10\%$ of the initial energy of the bob is dissipated as heat due to the friction of air. The speed of the bob when it reaches the lowermost point $B$ is ($g=10m{s}^{-2})$

• A. $3m{s}^{-1}$
• B. $4m{s}^{-1}$
• C. $5.3m{s}^{-1}$
• D. $6m{s}^{-1}$

Answer 1

The correct option is C $5.3m{s}^{-1}$

Energy converted$=95$% mgh

When v is velocity acquired at B, then

K.E$=\frac{95}{100}mgh$

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

$=\frac{1}{2}m{v}^2=\frac{95}{100}mgh$

$v=\sqrt{\frac{95}{100}\times 2gh}$

$v=\sqrt{\frac{19}{20}\times 2\times 9.8\times 1.8}=5.3m/s$