A 6 kg flower pot falls from a window ledge 10 meters above the ground.
a) What is the Kinetic Energy of the flower pot just as it reaches the ground?
b) What is the speed of the flower pot just before it strikes the ground?
KE at end=PE at start=mgh
V^2=2gh
5n
To answer both questions, we can make use of the principles of potential energy and kinetic energy. Potential energy refers to the energy an object possesses due to its position, while kinetic energy refers to the energy of an object in motion.
a) To find the kinetic energy of the flower pot just as it reaches the ground, we need to convert the potential energy it gained from falling into kinetic energy. The equation for potential energy is PE = m * g * h, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s² on Earth), and h is the height.
In this case, the mass of the flower pot is 6 kg, the height is 10 meters, and g is 9.8 m/s². Plugging these values into the equation, we get:
PE = 6 kg * 9.8 m/s² * 10 m
PE = 588 Joules
At the moment just before it strikes the ground, all potential energy is converted into kinetic energy. So the kinetic energy of the flower pot is equal to the potential energy, which is 588 Joules.
b) The equation for kinetic energy is KE = 1/2 * m * v², where KE is the kinetic energy, m is the mass, and v is the velocity (or speed).
To find the velocity of the flower pot just before it strikes the ground, we need to rearrange the equation. Let's solve for v:
KE = 1/2 * m * v²
2 * KE = m * v²
v² = (2 * KE) / m
v = √(2 * KE / m)
Using the kinetic energy we found in part a (588 Joules) and the mass of the flower pot (6 kg), we can plug them into the equation:
v = √(2 * 588 J / 6 kg)
v = √(196 J / kg)
v = √(196 m²/s²)
v = 14 m/s
Therefore, the speed of the flower pot just before it strikes the ground is 14 meters per second.