A 2kg stone is dropped from the top 20cm building. (a) what pe does it posses (b) at what height does its pe becomes equal to its ke c. What is its ke just before it hits the ground d. With what velocity does it reach the ground
To answer these questions, we need to understand the concepts of potential energy (PE) and kinetic energy (KE).
(a) The potential energy possessed by an object is given by the formula: PE = mgh, where m is the mass of the object (2kg in this case), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height (20cm or 0.2m). So, we can calculate the potential energy as follows:
PE = 2kg x 9.8 m/s² x 0.2m
PE = 3.92 Joules
Therefore, the stone possesses approximately 3.92 Joules of potential energy when it is dropped from the building's top.
(b) At what height does its potential energy become equal to its kinetic energy?
We know that at any given height, the potential energy is equal to the kinetic energy. So, we can set up the equation:
mgh = (1/2)mv²
Here, m represents the mass, g the acceleration due to gravity, h the height, and v the velocity. Since we're looking for the height, we can solve the equation for h:
h = (v² / (2g))
However, we need to know the velocity v at a particular height to calculate it precisely.
(c) What is its kinetic energy just before it hits the ground?
At the height just before the stone hits the ground, all its potential energy has been converted to kinetic energy. So, in this case, we can use the equation:
KE = mgh
Substituting the values:
KE = 2kg x 9.8 m/s² x 0.2m
KE = 0.784 Joules
Therefore, the kinetic energy just before it hits the ground is approximately 0.784 Joules.
(d) With what velocity does it reach the ground?
To calculate the velocity with which the stone reaches the ground, we can use the concept of conservation of energy. At the height of the ground, all potential energy has been converted to kinetic energy. So:
KE = mgh
Substituting the values:
KE = 2kg x 9.8 m/s² x 0.2m
KE = 0.784 Joules
Since KE = (1/2)mv², we can solve for v:
0.784 = (1/2) x 2kg x v²
v² = 0.784 m²/s²
v ≈ 0.886 m/s
Therefore, the stone reaches the ground with a velocity of approximately 0.886 m/s.