Formula of conservation of energy: **KE**_{i}** + U**_{i}** = KE**_{f}** + U**** _{f}** states that the sum of

**initial kinetic energy**(KE

_{i}) and

**initial potential energy**(U

_{i}) is equal to the sum of

**final kinetic energy**(KE

_{f}) and

**final potential energy**(U

_{f}).

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/s^{2})

Solution:

Given data:

Mass of a block, m = 10 kg

Initial velocity of a block, v_{i} = 0 m/s (Because, a block is at rest)

Initial height of a block, h_{i} = 40 m

Final velocity of a block, v_{f} = ?

Final height of a block, h_{f} = 10 m

Gravitational acceleration, g = 9.81 m/s^{2}

Using the formula of conservation of energy,

KE_{i} + U_{i} = KE_{f} + U_{f}

½ m v_{i}^{2} + m g h_{i} = ½ m v_{f}^{2} + m g h_{f}

½ v_{i}^{2} + g h_{i} = ½ v_{f}^{2} + g h_{f}

0 + (9.81 × 40) = ½ v_{f}^{2} + (9.81 × 10)

392.4 = ½ v_{f}^{2} + 98.1

½ v_{f}^{2} = 294.3

v_{f}^{2} = 588.6

v_{f} = 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/s^{2})

Solution:

Given data:

Mass of a basketball, m = 500 gm = 0.5 kg

Initial height of a basketball, h_{i} = 0.5 m

Final height of a basketball, h_{f} = 2.5 m

Initial velocity of a basketball, v_{i} = 0 m/s

Final velocity of a basketball, v_{f} = ?

Gravitational acceleration, g = 9.81 m/s^{2}

Using the formula of conservation of energy,

KE_{i} + U_{i} = KE_{f} + U_{f}

½ m v_{i}^{2} + m g h_{i} = ½ m v_{f}^{2} + m g h_{f}

½ v_{i}^{2} + g h_{i} = ½ v_{f}^{2} + g h_{f}

0 + (9.81 × 0.5) = ½ v_{f}^{2} + (9.81 × 2.5)

4.905 = ½ v_{f}^{2} + 24.525

½ v_{f}^{2} = 19.62

v_{f}^{2} = 39.24

v_{f} = 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/s^{2})

Solution:

Given data:

Mass of a stone, m = 15 kg

Initial velocity of a stone, v_{i} = 0 m/s (Because, a stone is at rest)

Initial height of a stone, h_{i} = 70 m

Final velocity of a stone, v_{f} = ?

Final height of a stone, h_{f} = 15 m

Gravitational acceleration, g = 9.81 m/s^{2}

Using the formula of conservation of energy,

KE_{i} + U_{i} = KE_{f} + U_{f}

½ m v_{i}^{2} + m g h_{i} = ½ m v_{f}^{2} + m g h_{f}

½ v_{i}^{2} + g h_{i} = ½ v_{f}^{2} + g h_{f}

0 + (9.81 × 70) = ½ v_{f}^{2} + (9.81 × 15)

686.7 = ½ v_{f}^{2} + 147.15

½ v_{f}^{2} = 539.55

v_{f}^{2} = 1079.1

v_{f} = 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/s^{2})

Solution:

Given data:

Mass of a kite, m = 20 gm = 0.02 kg

Initial height of a kite, h_{i} = 50 m

Initial velocity of a kite, v_{i} = 1 m/s

Final velocity of a kite, v_{f} = ?

Final height of a kite, h_{f} = 5 m

Gravitational acceleration, g = 9.81 m/s^{2}

Using the formula of conservation of energy,

KE_{i} + U_{i} = KE_{f} + U_{f}

½ m v_{i}^{2} + m g h_{i} = ½ m v_{f}^{2} + m g h_{f}

½ v_{i}^{2} + g h_{i} = ½ v_{f}^{2} + g h_{f}

(½ × 1^{2}) + (9.81 × 50) = ½ v_{f}^{2} + (9.81 × 5)

0.5 + 490.5 = ½ v_{f}^{2} + 49.05

491 = ½ v_{f}^{2} + 49.05

½ v_{f}^{2} = 441.95

v_{f}^{2} = 883.9

v_{f} = 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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**Related:**

- Thermal Energy Equation
- Potential Energy Formula
- Gravitational Potential Energy Formula
- Electric Potential Energy Formula
- Elastic Potential Energy Formula
- Kinetic Energy Formula
- Rotational Kinetic Energy Formula
- Electrical Energy Equation
- Mechanical Energy Formula
- Photon Energy Equation

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