krit.club logo

Energy - Mechanical Energy (Kinetic and Potential)

Grade 7ICSE

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

🔑Concepts

•

Energy is defined as the capacity to do work. The SI unit of energy is the Joule (JJ) and the CGS unit is the erg (1J=107 erg1 J = 10^7 \text{ erg}).

•

Mechanical Energy is the energy possessed by a body due to its state of rest or motion. It is the sum of Kinetic Energy and Potential Energy.

•

Kinetic Energy (K.E.K.E.) is the energy possessed by a body by virtue of its motion. Any object moving with a velocity vv possesses kinetic energy.

•

Potential Energy (P.E.P.E.) is the energy stored in a body due to its position (Gravitational Potential Energy) or its configuration (Elastic Potential Energy).

•

Gravitational Potential Energy depends on the mass of the body (mm), the acceleration due to gravity (gg), and the height (hh) above the reference level.

•

According to the Law of Conservation of Energy, energy can neither be created nor destroyed; it can only be transformed from one form to another. In a closed system, Total Mechanical Energy=K.E.+P.E.=constantTotal \, Mechanical \, Energy = K.E. + P.E. = \text{constant}.

📐Formulae

K.E.=12mv2K.E. = \frac{1}{2} m v^2

P.E.=mghP.E. = m g h

Total Mechanical Energy=K.E.+P.E.Total \, Mechanical \, Energy = K.E. + P.E.

1 Joule=1 kg m2s−21 \text{ Joule} = 1 \text{ kg m}^2 \text{s}^{-2}

💡Examples

Problem 1:

A ball of mass 0.5 kg0.5 \text{ kg} is kicked and moves with a velocity of 4 m/s4 \text{ m/s}. Calculate its Kinetic Energy.

Solution:

K.E.=12mv2=12×0.5×(4)2=0.25×16=4 JK.E. = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.5 \times (4)^2 = 0.25 \times 16 = 4 \text{ J}

Explanation:

Using the kinetic energy formula, we substitute the mass (m=0.5 kgm = 0.5 \text{ kg}) and velocity (v=4 m/sv = 4 \text{ m/s}) to find the energy in Joules.

Problem 2:

Find the gravitational potential energy of a stone of mass 2 kg2 \text{ kg} kept at a height of 10 m10 \text{ m} above the ground. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

P.E.=mgh=2×10×10=200 JP.E. = m g h = 2 \times 10 \times 10 = 200 \text{ J}

Explanation:

Potential energy is calculated by multiplying the mass of the object, the acceleration due to gravity, and the height above the ground.

Problem 3:

An object of mass 5 kg5 \text{ kg} is dropped from a height. If its velocity is 10 m/s10 \text{ m/s} just before hitting the ground, calculate its Kinetic Energy at that moment.

Solution:

K.E.=12×5×(10)2=12×5×100=250 JK.E. = \frac{1}{2} \times 5 \times (10)^2 = \frac{1}{2} \times 5 \times 100 = 250 \text{ J}

Explanation:

The energy of the falling object changes from potential energy at the top to kinetic energy at the bottom. At the moment of impact, the velocity is used to determine the K.E.K.E.