Question

At the low point in its swing, a pendulum bob with a mass of 0.2 kg has a velocity of 5m/s. How high will the bob swing above this low point?

Answer 1

For solving this task we will use energy conservation law:

There is a conversion of kinetic energy into potential. At the top point $E_{k} = 0$ but $E_{p} = \max$ , at the bottom point $E_{k} = \max$ but $E_{p} = 0$ , and $E_{k}$ at the bottom point is equal the $E_{p}$ at the top point.

$E_{k} = E_{p},$ where $E_{k} = \frac{mv^{2}}{2} -\text{kinetic energy},E_{p} = mgh - \text{potential energy}$

$m = 0.2kg$

$v = 5\frac{m}{s}$

$g = 10\frac{m}{s^2}$

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

$\frac{v^2}{2} = gh$

$h = \frac{v^2}{2g} = \frac{25}{20} = 1.25\text{metrs}$