# Conservation of energy formula

The conservation of energy formula, known as 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) in a system. This formula reflects the principle of energy conservation, highlighting that energy can be transferred or transformed but not created or destroyed. By applying this formula, one can analyze and predict changes in the distribution of kinetic and potential energy within a system.

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## Practice problems

### Problem #1

A 10 kg block, initially at rest, is located on the edge of a hill at a height of 40 m. The block falls down the hill. Calculate the velocity of the block when it is at a height of 10 m. Take the value of gravitational acceleration as 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

Applying the formula:

• 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 g basketball from a height of 0.5 m to a height of 2.5 m. If the initial velocity of the basketball is 0 m/s, calculate the velocity of the basketball when it reaches a height of 2.5 m. Take the value of gravitational acceleration as 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

Applying the formula:

• 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, initially at rest, is situated on the edge of a hill at a height of 70 m. The stone falls down the hill. What is the velocity of the stone when it is at a height of 15 m before hitting the ground? Take the value of gravitational acceleration as 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

Applying the formula:

• 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 g kite is flying in the sky at a height of 50 m above the ground and is descending with a velocity of 1 m/s. Calculate the velocity of the kite when it is at a height of 5 m. Take the value of gravitational acceleration as 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

Applying the formula:

• 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.