A pendulum bob with a mass of 0.4 kg is released from a position in which the bob is 20 cm above the low point in its swing. What’s the speed of the bob as it passes through the low point in its swing?
To find the speed of the pendulum bob as it passes through the low point in its swing, we can use the principle of conservation of energy.
1. First, let's calculate the potential energy of the pendulum bob at its initial position (20 cm above the low point). The potential energy of an object at height h, mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, h = 0.20 m (converting 20 cm to meters).
Potential Energy = m * g * h = 0.4 kg * 9.8 m/s^2 * 0.20 m = 0.784 Joules
2. When the pendulum bob is at the low point of its swing, all the potential energy is converted into kinetic energy. The formula for kinetic energy is given by (1/2) * m * v^2, where m is the mass of the object and v is its velocity.
0.784 Joules = (1/2) * 0.4 kg * v^2
3. Now we can solve for v.
0.784 Joules = 0.2 kg * v^2
0.784 Joules / 0.2 kg = v^2
3.92 m^2/s^2 = v^2
Taking the square root of both sides, we find:
v ≈ 1.98 m/s
Therefore, the speed of the pendulum bob as it passes through the low point in its swing is approximately 1.98 m/s.