Question

A pendulum of mass 240 g undergoes simple harmonic motion when acted upon by a force of 14 N. The pendulum crosses the point of equilibrium at a speed of 6 m · s−1. What is the energy (in J) of the pendulum at the center of the oscillation?

Answer 1

Answer:

4.3J

Explanation:

The energy associated with a simple pendulum is the potential energy and the kinetic energy. At rest the kinetic energy is zero while the potential energy is maximize, also at the center of the oscillation, the potential energy is zero while the kinetic energy is maximize.

Hence from the question given, we can conclude that the only associated energy is the Kinetic energy which is expressed  as

$Kinetic Energy = \frac{1}{2}mv^{2}$

since mass,m=240g=0.24kg,

Velocity V=6m/s

If we substitute values we arrive at

$Kinetic Energy = \frac{1}{2}mv^{2} \\Kinetic Energy = \frac{1}{2} *0.24*6^{2}\\Kinetic Energy =4.3J$