Problem 3

Since energy is conserved, the total of the kinetic energy ( ${E}_{K}$ ) and gravitational potential energy ( ${E}_{GP}$ ) combined is the same at every instance.

Consider the instance before the pendulum is dropped. The pendulum is not moving, so ${E}_{K}=\dfrac{1}{2}mv^2=0$ . However there is gravitational potential energy - if we take the lowest point of the pendulum to be the ground then ${E}_{GP}=mgh=m \times 10 \times (0.1-0.1\sin30^\circ)=\dfrac{m}{2}$ . Therefore the total energy is $\dfrac{m}{2}$ .

Consider the instance when the pendulum is at its lowest point. ${E}_{K}=\dfrac{1}{2}mv^2$ , but ${E}_{GP}=mgh=m \times 10 \times 0=0$ . Therefore the total energy is $\dfrac{1}{2}mv^2$ .

$\dfrac{m}{2}=\dfrac{1}{2}mv^2 \, \Rightarrow \, v^2=1 \, \Rightarrow \, v=\large\color{#20A900}{\boxed{1}}$

mgl/2=1/2mv² v=√lg=1m/s