A frictionless pendulum with a mass of 0.4 kg and a length of 2.1 m starts at point A, at an angle θ of 60°. As it swings downward, it passes through point B, which is 30 degrees from equilibrium. The kinetic energy of the pendulum at point B is 2.98 J.
The diagrammatic illustration of the question can be seen in the image attached below.
Applying the conservation of energy;
- change in kinetic energy = change in potential energy
[tex]\mathbf{K_B -K_A = \Delta P_{BA}}[/tex]
Since we are to determine the kinetic energy of the pendulum at point B.
Let make [tex]\mathbf{K_B }[/tex] is the subject of the formula:
[tex]\mathbf{K_B = \Delta P_{BA}+ K_A}[/tex]
[tex]\mathbf{K_B = \Delta P_{BA}+ \dfrac{1}{2}mV_A^2}[/tex]
where;
- [tex]\mathbf{V_A = 0}[/tex]
[tex]\mathbf{K_B = \Delta P_{BA}+ \dfrac{1}{2}m(0)^2}[/tex]
[tex]\mathbf{K_B = \Delta P_{BA}}[/tex]
[tex]\mathbf{K_B =m \times g \Delta h }[/tex]
From the diagram,
Δ h = (2.1 cos 30° - 2.1 cos 60°)
Δ h = 1.81 - 1.05
Δ h = 0.76
[tex]\mathbf{K_B =m \times g \Delta h }[/tex]
[tex]\mathbf{K_B =0.4 \times 9.8 \times 0.76 }[/tex]
[tex]\mathbf{K_B =2.98 \ J }[/tex]
Therefore, we can conclude that the kinetic energy of the pendulum at point B is 2.98 J
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