In compressed state, a spring 3.00 meters long hanging from the ceiling has a potential energy of 2.0 joules. If the entire potential energy is used to release a ball of mass 1.2 kilograms vertically downward, then what is the kinetic energy of the ball when it reaches the ground?
To find the kinetic energy of the ball when it reaches the ground, we need to consider the conservation of energy.
The potential energy stored in the compressed spring is converted into the kinetic energy of the ball as it falls. The total mechanical energy remains constant throughout the process.
The potential energy (PE) of the spring is given as 2.0 joules. This potential energy will be fully converted into kinetic energy (KE) as the ball falls.
The formula to calculate potential energy is given by:
PE = m * g * h
Where:
m is the mass of the ball (1.2 kilograms),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
h is the height through which the ball falls.
We can rearrange the formula to solve for height (h):
h = PE / (m * g)
Substituting the given values:
h = 2.0 joules / (1.2 kilograms * 9.8 m/s^2)
= 2.0 joules / 11.76 joules/kg*m^2
≈ 0.17 meters
So, the ball falls through a height of approximately 0.17 meters.
Now, let's calculate the kinetic energy (KE) of the ball when it reaches the ground.
The formula for calculating kinetic energy is:
KE = 0.5 * m * v^2
Where:
m is the mass of the ball (1.2 kilograms),
v is the velocity of the ball.
To calculate the velocity (v), we can use the equation of motion:
v^2 = u^2 + 2 * a * s
Where:
u is the initial velocity of the ball (0 m/s),
a is the acceleration due to gravity (9.8 m/s^2),
s is the distance through which the ball falls (0.17 meters).
Substituting the given values:
v^2 = 0^2 + 2 * 9.8 m/s^2 * 0.17 meters
= 0 + 3.332 m^2/s^2
≈ 3.33 m^2/s^2
Taking the square root of both sides:
v ≈ √(3.33 m^2/s^2)
≈ 1.83 m/s
Now, we can calculate the kinetic energy (KE) using the formula:
KE = 0.5 * m * v^2
Substituting the given values:
KE = 0.5 * 1.2 kilograms * (1.83 m/s)^2
= 0.5 * 1.2 kilograms * 3.35 m^2/s^2
= 2.01 joules
Therefore, the kinetic energy of the ball when it reaches the ground is approximately 2.01 joules.