A 4.50 kg demonic cat is dropped from a 25.0 m building.
What will its final velocity be if all of the gravitational PE is converted to KE?
Round to 3 SFs. Do NOT include the unit (m/s).
1/2 m v^2 = m g h ... v = √(2 g h) = √(2 * 9.81 * 25.0) m/s
To find the final velocity (v) of the demonic cat after falling from a building, we can use the law of conservation of energy. In this case, all the gravitational potential energy (PE) is converted to kinetic energy (KE). The formula for gravitational potential energy is:
PE = m * g * h
Where:
m = mass of the cat = 4.50 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the building = 25.0 m
Let's calculate the gravitational potential energy:
PE = 4.50 kg * 9.8 m/s^2 * 25.0 m
PE = 1,102.5 J
Since all the gravitational potential energy is converted to kinetic energy, the final kinetic energy (KE) will be equal to the initial potential energy (PE). The formula for kinetic energy is:
KE = (1/2) * m * v^2
Where:
m = mass of the cat = 4.50 kg
v = final velocity of the cat
Setting the initial potential energy equal to the final kinetic energy:
PE = KE
1,102.5 J = (1/2) * 4.50 kg * v^2
Simplifying the equation:
v^2 = (2 * 1,102.5 J) / 4.50 kg
v^2 = 2,205 J / 4.50 kg
v^2 = 490 m^2/s^2
Taking the square root of both sides to solve for v:
v = √(490 m^2/s^2)
v ≈ 22.1 m/s
Rounding to 3 significant figures, the final velocity of the demonic cat will be approximately 22.1 m/s.