A stone of mass 1 kg and initially at rest is dropped from a tower of height 40 m. When it reaches a height of 10 m from the ground level, what will be the values of its potential energy (PE) and kinetic energy (KE)? (Acceleration due to gravity is 10 m/s²)
PE = 300 J, KE = 100 J
PE = 200 J, KE = 200 J
PE = 100 J, KE = 300 J
PE = 100 J, KE = 200 J
Answer is Right!
Answer is Wrong!
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UPSC Geoscientist – 2021
Initial state (at height 40 m, at rest):
Initial PE = mgh₁ = 1 kg * 10 m/s² * 40 m = 400 J
Initial KE = ½ mv₁² = ½ * 1 kg * (0 m/s)² = 0 J
Total Energy = Initial PE + Initial KE = 400 J + 0 J = 400 J.
Final state (at height 10 m):
Final PE = mgh₂ = 1 kg * 10 m/s² * 10 m = 100 J.
Since Total Energy is conserved:
Total Energy = Final PE + Final KE
400 J = 100 J + Final KE
Final KE = 400 J – 100 J = 300 J.
So, at a height of 10 m, PE = 100 J and KE = 300 J.