If an object of mass 2 kg is dropped from a height of 10 metres, what will be the ratio of its potential energy and kinetic energy at the height of 5 metres (g = 10m/sec² )

Question

Question

This question was previously asked in

SSC CHSL Exam 2024 Tier-I Official Paper (Held On: 09 Jul, 2024 Shift 2)

Answer 1

The correct answer is 1 : 1

Key Points

Potential Energy (PE) at a height of 10 meters is given by the formula: PE = m * g * h
For m = 2 kg, g = 10 m/s², and h = 10 m, PE = 2 * 10 * 10 = 200 Joules
Potential Energy at a height of 5 meters: PE = m * g * h = 2 * 10 * 5 = 100 Joules
Kinetic Energy (KE) at a height of 5 meters can be found using the conservation of energy principle: Total Energy at 10 meters = Total Energy at 5 meters
Total Energy = PE at 10 meters = 200 Joules
At 5 meters, Total Energy = PE (5 meters) + KE (5 meters) = 100 + KE
Therefore, KE at 5 meters = 200 - 100 = 100 Joules
The ratio of Potential Energy to Kinetic Energy at 5 meters is PE : KE = 100 : 100 = 1 : 1

Additional Information

Potential Energy is the energy possessed by an object due to its position relative to a reference point.
Kinetic Energy is the energy possessed by an object due to its motion.
The law of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another.
In free fall, the sum of potential energy and kinetic energy remains constant if air resistance is neglected.