Conservation of Energy Formula | Problems (And Solutions)

Conservation of Energy Formula

Formula of conservation of energy: KEi + Ui = KEf + Uf states that the sum of initial kinetic energy (KEi) and initial potential energy (Ui) is equal to the sum of final kinetic energy (KEf) and final potential energy (Uf).

Let’s solve some problems based on this formula, so you’ll get a clear idea.

Conservation of Energy Practice Problems

Problem 1: A 10 kg block which is at rest, sitting on the edge of a hill at a height of 40 m, falls down from a hill. Calculate the velocity of a block at a height of 10 m. (Take the value of gravitational acceleration, g = 9.81 m/s2)

Solution:

Given data:
Mass of a block, m = 10 kg
Initial velocity of a block, vi = 0 m/s (Because, a block is at rest)
Initial height of a block, hi = 40 m
Final velocity of a block, vf = ?
Final height of a block, hf = 10 m
Gravitational acceleration, g = 9.81 m/s2

Using the formula of conservation of energy,
KEi + Ui = KEf + Uf
½ m vi2 + m g hi = ½ m vf2 + m g hf
½ vi2 + g hi = ½ vf2 + g hf
0 + (9.81 × 40) = ½ vf2 + (9.81 × 10)
392.4 = ½ vf2 + 98.1
½ vf2 = 294.3
vf2 = 588.6
vf = 24.26 m/s

Therefore, the velocity of a block at a height of 10 m is 24.26 m/s.


Problem 2: A basketball player throws a 500 gm basketball from a 0.5 m height to a 2.5 m height. If the initial velocity of a basketball is 0 m/s, then calculate the velocity of a basketball at a height of 2.5 m. (Take the value of gravitational acceleration, g = 9.81 m/s2)

Solution:

Given data:
Mass of a basketball, m = 500 gm = 0.5 kg
Initial height of a basketball, hi = 0.5 m
Final height of a basketball, hf = 2.5 m
Initial velocity of a basketball, vi = 0 m/s
Final velocity of a basketball, vf = ?
Gravitational acceleration, g = 9.81 m/s2

Using the formula of conservation of energy,
KEi + Ui = KEf + Uf
½ m vi2 + m g hi = ½ m vf2 + m g hf
½ vi2 + g hi = ½ vf2 + g hf
0 + (9.81 × 0.5) = ½ vf2 + (9.81 × 2.5)
4.905 = ½ vf2 + 24.525
½ vf2 = 19.62
vf2 = 39.24
vf = 6.26 m/s

Therefore, the velocity of a basketball at a height of 2.5 m is 6.26 m/s.


Problem 3: A 15 kg stone at rest, sitting on the edge of a hill at a height of 70 m, falls down from a hill. What is the velocity of a stone at a height of 15 m before it hits the ground? (Take the value of gravitational acceleration, g = 9.81 m/s2)

Solution:

Given data:
Mass of a stone, m = 15 kg
Initial velocity of a stone, vi = 0 m/s (Because, a stone is at rest)
Initial height of a stone, hi = 70 m
Final velocity of a stone, vf = ?
Final height of a stone, hf = 15 m
Gravitational acceleration, g = 9.81 m/s2

Using the formula of conservation of energy,
KEi + Ui = KEf + Uf
½ m vi2 + m g hi = ½ m vf2 + m g hf
½ vi2 + g hi = ½ vf2 + g hf
0 + (9.81 × 70) = ½ vf2 + (9.81 × 15)
686.7 = ½ vf2 + 147.15
½ vf2 = 539.55
vf2 = 1079.1
vf = 32.84 m/s

Therefore, the velocity of a stone at a height of 15 m is 32.84 m/s.


Problem 4: A 20 gm kite is flying in the sky at a height of 50 m above the ground and is coming down with the velocity of 1 m/s. Calculate the velocity of a kite at a height of 5 m. (Take the value of gravitational acceleration, g = 9.81 m/s2)

Solution:

Given data:
Mass of a kite, m = 20 gm = 0.02 kg
Initial height of a kite, hi = 50 m
Initial velocity of a kite, vi = 1 m/s
Final velocity of a kite, vf = ?
Final height of a kite, hf = 5 m
Gravitational acceleration, g = 9.81 m/s2

Using the formula of conservation of energy,
KEi + Ui = KEf + Uf
½ m vi2 + m g hi = ½ m vf2 + m g hf
½ vi2 + g hi = ½ vf2 + g hf
(½ × 12) + (9.81 × 50) = ½ vf2 + (9.81 × 5)
0.5 + 490.5 = ½ vf2 + 49.05
491 = ½ vf2 + 49.05
½ vf2 = 441.95
vf2 = 883.9
vf = 29.73 m/s

Therefore, the velocity of a kite at a height of 5 m is 29.73 m/s.

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