The given problem can be exemplified in the following diagram:
To determine the velocity when the mass is located at point 2 we will consider that the change in gravitational potential energy from 1 to 2 is equal to the kinetic energy at 2, therefore, we have:
[tex]\Delta U=K[/tex]Where:
[tex]\begin{gathered} \Delta U=\text{ change in gravitational potential energy} \\ K=\text{ kinetic energy} \end{gathered}[/tex]Now, we use the following formula for the gravitational potential energy:
[tex]\Delta U=mgh[/tex]Where:
[tex]\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \\ h=\text{ change in height} \end{gathered}[/tex]The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]Now, we substitute the formulas:
[tex]mgh=\frac{1}{2}mv^2[/tex]We can cancel out the mass "m":
[tex]gh=\frac{1}{2}v^2[/tex]Now, we solve for the velocity by multiplying both sides by 2:
[tex]2gh=v^2[/tex]Now, we take the square root to both sides:
[tex]\sqrt[]{2gh}=v[/tex]Now, we determine the height "h" using the following triangle:
We notice that the adjacent side of the triangle plus the height "h" is equal to the length of the pendulum, therefore, we have the next relationship:
[tex]x+h=2[/tex]We solve for "h" by subtracting "x" from both sides:
[tex]h=2-x[/tex]Now, we determine the value of "x" by using the function cosine:
[tex]\cos 33=\frac{x}{2}[/tex]Now, we multiply both sides by 2:
[tex]2\cos 33=x[/tex]Now, we substitute the value of "x":
[tex]h=2-2\cos 33[/tex]Solving the operations:
[tex]h=0.32[/tex]Therefore, the change in height is 0.32 meters. Substituting in the formula for the velocity we get:
[tex]\sqrt[]{2(9.8\frac{m}{s^2})(0.32m)}=v[/tex]Solving the operations:
[tex]2.5\frac{m}{s}=v[/tex]Therefore, the velocity of the mass is 2.5 m/s.
