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Physics: Forces and Energy - Kinetic and Potential Energy

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Kinetic Energy (EkE_k) is the energy possessed by an object due to its motion. Any object with mass (mm) moving at a specific velocity (vv) has kinetic energy.

Gravitational Potential Energy (EpE_p) is the energy stored in an object because of its position in a gravitational field, usually relative to the ground. It depends on mass, the gravitational field strength (g9.8 m/s2g \approx 9.8 \text{ m/s}^2), and height (hh).

The Law of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one form to another. In a frictionless system, the total mechanical energy remains constant: Etotal=Ek+EpE_{total} = E_k + E_p.

The standard unit for energy is the Joule (JJ), where 1 J=1 kgm2/s21 \text{ J} = 1 \text{ kg} \cdot \text{m}^2/\text{s}^2.

Work Done is defined as the transfer of energy. When work is done on an object, it gains energy: W=ΔEW = \Delta E.

📐Formulae

Ek=12mv2E_k = \frac{1}{2}mv^2

Ep=mghE_p = mgh

Etotal=Ek+EpE_{total} = E_k + E_p

v=2Ekmv = \sqrt{\frac{2E_k}{m}}

💡Examples

Problem 1:

A 2 kg2\text{ kg} book is lifted to a shelf that is 3 m3\text{ m} high. Calculate the Gravitational Potential Energy (EpE_p) gained by the book. (Assume g=9.8 m/s2g = 9.8\text{ m/s}^2)

Solution:

Ep=mgh=2 kg×9.8 m/s2×3 m=58.8 JE_p = mgh = 2\text{ kg} \times 9.8\text{ m/s}^2 \times 3\text{ m} = 58.8\text{ J}

Explanation:

The potential energy is found by multiplying the mass, the acceleration due to gravity, and the vertical height.

Problem 2:

A toy car with a mass of 0.5 kg0.5\text{ kg} is moving at a velocity of 4 m/s4\text{ m/s}. Determine its Kinetic Energy (EkE_k).

Solution:

Ek=12mv2=12×0.5 kg×(4 m/s)2=0.25×16=4 JE_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5\text{ kg} \times (4\text{ m/s})^2 = 0.25 \times 16 = 4\text{ J}

Explanation:

To find kinetic energy, square the velocity first, then multiply by the mass and divide by two.

Problem 3:

A ball of mass 0.1 kg0.1\text{ kg} is dropped from a height of 5 m5\text{ m}. Ignoring air resistance, what is its velocity (vv) just before it hits the ground?

Solution:

mgh=12mv2    gh=12v2    v=2gh=2×9.8×5=989.9 m/smgh = \frac{1}{2}mv^2 \implies gh = \frac{1}{2}v^2 \implies v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.9\text{ m/s}

Explanation:

Using the Conservation of Energy, the initial potential energy at the top is converted entirely into kinetic energy at the bottom.

Kinetic and Potential Energy - Revision Notes & Key Formulas | IB Grade 8 Science