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Work, Energy and Simple Machines - Forms of Energy

Grade 9CBSE

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

🔑Concepts

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Energy is defined as the capacity to do work. The SI unit of energy is the Joule (JJ), where 1J=1Nm1 J = 1 Nm.

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Kinetic Energy (EkE_k) is the energy possessed by an object by virtue of its motion. It is directly proportional to the mass of the object and the square of its velocity.

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Potential Energy (EpE_p) is the energy stored in an object due to its position or change in shape. Gravitational potential energy depends on mass (mm), acceleration due to gravity (gg), and height (hh) above a reference point.

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The Law of Conservation of Energy states that energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system remains constant.

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Mechanical Energy is the sum total of kinetic energy and potential energy: ME=Ek+EpME = E_k + E_p.

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The commercial unit of energy is kilowatt-hour (kWhkWh). 1kWh1 kWh is the energy consumed by an electrical appliance of power 1kW1 kW in one hour.

📐Formulae

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

Ep=mghE_p = mgh

W=ΔEk=12m(v2−u2)W = \Delta E_k = \frac{1}{2}m(v^2 - u^2)

P=EtP = \frac{E}{t}

1 kWh=3.6×106 J1 \text{ kWh} = 3.6 \times 10^6 \text{ J}

💡Examples

Problem 1:

An object of mass 15 kg15 \text{ kg} is moving with a uniform velocity of 4 m/s4 \text{ m/s}. What is the kinetic energy possessed by the object?

Solution:

Given: m=15 kgm = 15 \text{ kg}, v=4 m/sv = 4 \text{ m/s}. Using the formula: Ek=12mv2E_k = \frac{1}{2}mv^2 Ek=12×15×(4)2E_k = \frac{1}{2} \times 15 \times (4)^2 Ek=12×15×16E_k = \frac{1}{2} \times 15 \times 16 Ek=15×8=120 JE_k = 15 \times 8 = 120 \text{ J}

Explanation:

The kinetic energy is calculated by substituting the mass and the square of the velocity into the standard kinetic energy equation.

Problem 2:

Find the energy possessed by an object of mass 10 kg10 \text{ kg} when it is at a height of 6 m6 \text{ m} above the ground. (Take g=9.8 m/s2g = 9.8 \text{ m/s}^2)

Solution:

Given: m=10 kgm = 10 \text{ kg}, g=9.8 m/s2g = 9.8 \text{ m/s}^2, h=6 mh = 6 \text{ m}. Using the formula: Ep=mghE_p = mgh Ep=10×9.8×6E_p = 10 \times 9.8 \times 6 Ep=98×6=588 JE_p = 98 \times 6 = 588 \text{ J}

Explanation:

Potential energy is calculated based on the work done against gravity to lift the mass to the specified height.

Problem 3:

An electric meter in a house shows a reading of 2350 kWh2350 \text{ kWh} at the start of the month and 2580 kWh2580 \text{ kWh} at the end. Calculate the energy consumed in Joules.

Solution:

First, find the energy consumed in kWhkWh using subtraction: 2580−2350230\begin{array}{r} 2580 \\ - 2350 \\ \hline 230 \end{array} Energy consumed = 230 kWh230 \text{ kWh}. To convert to Joules: E=230×3.6×106 JE = 230 \times 3.6 \times 10^6 \text{ J} E=828×106 J=8.28×108 JE = 828 \times 10^6 \text{ J} = 8.28 \times 10^8 \text{ J}

Explanation:

The total units consumed is the difference between final and initial meter readings. This value is then multiplied by the conversion factor for 1 kWh1 \text{ kWh} to reach the result in Joules.